Ionic Equilibrium
153 Questions
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 27th July Evening Shift
The plot of $\mathrm{pH}$-metric titration of weak base $\mathrm{NH}_{4} \mathrm{OH}$ vs strong acid HCl looks like :
A.
B.
C.
D.
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 26th July Evening Shift
Class XII students were asked to prepare one litre of buffer solution of $\mathrm{pH} \,8.26$ by their Chemistry teacher: The amount of ammonium chloride to be dissolved by the student in $0.2\, \mathrm{M}$ ammonia solution to make one litre of the buffer is :
(Given: $\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_{3}\right)=4.74$
Molar mass of $\mathrm{NH}_{3}=17 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar mass of $\mathrm{NH}_{4} \mathrm{Cl}=53.5 \mathrm{~g} \mathrm{~mol}^{-1}$ )
A.
53.5 g
B.
72.3 g
C.
107.0 g
D.
126.0 g
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 25th July Evening Shift
${K_{{a_1}}}$, ${K_{{a_2}}}$ and ${K_{{a_3}}}$ are the respective ionization constants for the following reactions (a), (b) and (c).
(a) ${H_2}{C_2}{O_4} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H^ + } + H{C_2}O_4^ - $
(b) $H{C_2}O_4^ - \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H^ + } + {C_2}O_4^{2 - }$
(c) ${H_2}{C_2}O_4^{} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} 2{H^ + } + {C_2}O_4^{2 - }$
The relationship between ${K_{{a_1}}}$, ${K_{{a_2}}}$ and ${K_{{a_3}}}$ is given as :
A.
${K_{{a_3}}}$ $=$ ${K_{{a_1}}}$ $+$ ${K_{{a_2}}}$
B.
${K_{{a_3}}}$ $=$ ${K_{{a_1}}}$ $-$ ${K_{{a_2}}}$
C.
${K_{{a_3}}}$ $=$ ${K_{{a_1}}}$ $/$ ${K_{{a_2}}}$
D.
${K_{{a_3}}}$ $=$ ${K_{{a_1}}}$ $\times$ ${K_{{a_2}}}$
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 25th July Morning Shift
$20 \mathrm{~mL}$ of $0.1\, \mathrm{M} \,\mathrm{NH}_{4} \mathrm{OH}$ is mixed with $40 \mathrm{~mL}$ of $0.05 \mathrm{M} \mathrm{HCl}$. The $\mathrm{pH}$ of the mixture is nearest to :
(Given : $\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=1 \times 10^{-5}, \log 2=0.30, \log 3=0.48, \log 5=0.69, \log 7=0.84, \log 11= 1.04)$
A.
3.2
B.
4.2
C.
5.2
D.
6.2
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 29th June Morning Shift
The solubility of AgCl will be maximum in which of the following?
A.
0.01 M KCl
B.
0.01 M HCl
C.
0.01 M AgNO3
D.
Deionised water
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 28th June Evening Shift
A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of ${{[C{H_3}C{H_2}CO{O^ - }]} \over {[C{H_3}C{H_2}COOH]}}$ required to make buffer is ___________.
Given : ${K_a}(C{H_3}C{H_2}COOH) = 1.3 \times {10^{ - 5}}$
A.
0.03
B.
0.13
C.
0.23
D.
0.33
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 25th June Evening Shift
The Ksp for bismuth sulphide (Bi2S3) is 1.08 $\times$ 10$-$73. The solubility of Bi2S3 in mol L$-$1 at 298 K is :
A.
1.0 $\times$ 10$-$15
B.
2.7 $\times$ 10$-$12
C.
3.2 $\times$ 10$-$10
D.
4.2 $\times$ 10$-$8
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 25th June Evening Shift
Given below are two statements one is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : The amphoteric nature of water is explained by using Lewis acid/base concept.
Reason R : Water acts as an acid with NH3 and as a base with H2S.
In the light of the above statements choose the correct answer from the options given below :
A.
Both A and R are true and R is the correct explanation of A.
B.
Both A and R are true but R is NOT the correct explanation of A.
C.
A is true but R is false.
D.
A is false but R is true.
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 29th July Morning Shift
If the solubility product of PbS is 8 $\times$ 10$-$28, then the solubility of PbS in pure water at 298 K is x $\times$ 10$-$16 mol L$-$1. The value of x is __________. (Nearest Integer)
[Given : $\sqrt2$ = 1.41]
Correct Answer: 282
Explanation:
$\mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2}$
$\mathrm{S}=\sqrt{K_{s p}}=\sqrt{8 \times 10^{-28}}=2 \sqrt{2} \times 10^{-14}$
$=2.82 \times 10^{-14}$
$=282 \times 10^{-16}$
$ \therefore $ Ans. 282
$\mathrm{S}=\sqrt{K_{s p}}=\sqrt{8 \times 10^{-28}}=2 \sqrt{2} \times 10^{-14}$
$=2.82 \times 10^{-14}$
$=282 \times 10^{-16}$
$ \therefore $ Ans. 282
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th July Morning Shift
$\mathrm{K}_{\mathrm{a}}$ for butyric acid $\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{COOH}\right)$ is $2 \times 10^{-5}$. The $\mathrm{pH}$ of $0.2 \,\mathrm{M}$ solution of butyric acid is __________ $\times 10^{-1}$. (Nearest integer)
[Given $\log 2=0.30$]
Correct Answer: 27
Explanation:
$\mathrm{K}_{\mathrm{a}}$ of Butyric acid $\Rightarrow 2 \times 10^{-5} \,\mathrm{PKa}=4.7$
$\mathrm{pH}$ of $0.2 \mathrm{M}$ solution,
$ \mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \log \mathrm{C} $
$ \begin{aligned} &=\frac{1}{2}(4 \cdot 7) - \frac{1}{2} \log (0.2) \\\\ &=2.35+0.35=2.7 \end{aligned} $
$ \mathrm{pH}=27 \times 10^{-1} $
$\mathrm{pH}$ of $0.2 \mathrm{M}$ solution,
$ \mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \log \mathrm{C} $
$ \begin{aligned} &=\frac{1}{2}(4 \cdot 7) - \frac{1}{2} \log (0.2) \\\\ &=2.35+0.35=2.7 \end{aligned} $
$ \mathrm{pH}=27 \times 10^{-1} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th July Morning Shift
At $310 \mathrm{~K}$, the solubility of $\mathrm{CaF}_{2}$ in water is $2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$. The solubility product of $\mathrm{CaF}_{2}$ is ____________ $\times 10^{-8}(\mathrm{~mol} / \mathrm{L})^{3}$. (Give molar mass : $\mathrm{CaF}_{2}=78 \mathrm{~g} \mathrm{~mol}^{-1}$)
Correct Answer: 0
Explanation:
$\mathrm{CaF}_{2} \stackrel{\mathrm{s}}{\rightleftharpoons} \underset{ \mathrm{s}}{\mathrm{Ca}^{2+}}+\underset{2 \mathrm{~s}}{2 \mathrm{~F}^{-}}$
$ \begin{aligned} \mathrm{K}_{\mathrm{sp}} &=\mathrm{s}(2 \mathrm{~s})^{2} \\\\ &=4 \mathrm{~s}^{3} \end{aligned} $
Solubility $(\mathrm{s})=2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$
$=\frac{2 \cdot 34 \times 10^{-3} \times 10}{78}$ mole $/$ lit
$=3 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}$
$\therefore \mathrm{K}_{\mathrm{sp}}=4 \times\left(3 \times 10^{-4}\right)^{3}$
$ \begin{aligned} &=108 \times 10^{-12} \\\\ &=0.0108 \times 10^{-8}(\mathrm{~mole} / \mathrm{lit})^{3} \end{aligned} $
$ \begin{aligned} & \therefore x \approx 0 \end{aligned} $
$ \begin{aligned} \mathrm{K}_{\mathrm{sp}} &=\mathrm{s}(2 \mathrm{~s})^{2} \\\\ &=4 \mathrm{~s}^{3} \end{aligned} $
Solubility $(\mathrm{s})=2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$
$=\frac{2 \cdot 34 \times 10^{-3} \times 10}{78}$ mole $/$ lit
$=3 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}$
$\therefore \mathrm{K}_{\mathrm{sp}}=4 \times\left(3 \times 10^{-4}\right)^{3}$
$ \begin{aligned} &=108 \times 10^{-12} \\\\ &=0.0108 \times 10^{-8}(\mathrm{~mole} / \mathrm{lit})^{3} \end{aligned} $
$ \begin{aligned} & \therefore x \approx 0 \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th July Morning Shift
In the titration of $\mathrm{KMnO}_{4}$ and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is ___________.
Correct Answer: 1
Explanation:
$16 \mathrm{H}^{+}+2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 10 \mathrm{CO}_{2}+2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}$
During titration of oxalic acid by $\mathrm{KMnO}_{4}$, oxalic acid converts into $\mathrm{CO}_{2}$.
$\therefore$ Change in oxidation state of carbon $=1$
During titration of oxalic acid by $\mathrm{KMnO}_{4}$, oxalic acid converts into $\mathrm{CO}_{2}$.
$\therefore$ Change in oxidation state of carbon $=1$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th June Morning Shift
The solubility product of a sparingly soluble salt A2X3 is 1.1 $\times$ 10$-$23. If specific conductance of the solution is 3 $\times$ 10$-$5 S m$-$1, the limiting molar conductivity of the solution is $x \,\times$ 10$-$3 S m2 mol$-$1. The value of x is ___________.
Correct Answer: 3
Explanation:
$A_{2} X_{3} \rightleftharpoons \underset{2S}{2 \mathrm{~A}}+\underset{3S}{3 \mathrm{X}}$
$ \begin{aligned} &\mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^{2}(3 s)^{3}=1.1 \times 10^{-23} \\\\ &\mathrm{~S} \approx 10^{-5} \end{aligned} $
For sparingly soluble salts
$ \begin{aligned} \wedge_{m} &=\wedge_{m}^{0} \\\\ \wedge_{m} &=\frac{\mathrm{k}}{\mathrm{S} \times 10^{3}} \\\\ &=\frac{3 \times 10^{-5}}{10^{-5}} \times 10^{-3} \\\\ &=3 \times 10^{-3} ~ \mathrm{Sm}^{2} \mathrm{~mol}^{-1} \end{aligned} $
$ \begin{aligned} &\mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^{2}(3 s)^{3}=1.1 \times 10^{-23} \\\\ &\mathrm{~S} \approx 10^{-5} \end{aligned} $
For sparingly soluble salts
$ \begin{aligned} \wedge_{m} &=\wedge_{m}^{0} \\\\ \wedge_{m} &=\frac{\mathrm{k}}{\mathrm{S} \times 10^{3}} \\\\ &=\frac{3 \times 10^{-5}}{10^{-5}} \times 10^{-3} \\\\ &=3 \times 10^{-3} ~ \mathrm{Sm}^{2} \mathrm{~mol}^{-1} \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th June Evening Shift
pH value of 0.001 M NaOH solution is ____________.
Correct Answer: 11
Explanation:
$
\begin{aligned}
&{\left[\mathrm{OH}^{-}\right]=0.001=10^{-3} \mathrm{M}} \\\\
&{\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}} \\\\
&\quad\left[\mathrm{H}^{+}\right]=10^{-11} \\\\
&\begin{aligned}
\mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\\\
=&-\log \left(10^{-11}\right) \\\\
\mathrm{pH} &=11
\end{aligned}
\end{aligned}
$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th June Morning Shift
50 mL of 0.1 M CH3COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be _____________ $\times$ 10$-$2. (Nearest integer)
(Given : pKa (CH3COOH) = 4.76)
log 2 = 0.30
log 3 = 0.48
log 5 = 0.69
log 7 = 0.84
log 11 = 1.04
Correct Answer: 476
Explanation:
CH3COOH + NaOH $\to$ CH3COONa + H2O
After adding 25 ml of NaOH volume of mixture = 50 + 25 = 75 ml
Initially,
Number of millimole of NaOH = 25 $\times$ 0.1 = 2.5 mm
Number of millimole of CH3COOH = 50 $\times$ 0.1 = 5 mm
After nutrilisation,
Millimole of NaOH = 0
Millimole of CH3COOH = 5 $-$ 2.5 = 2.5 mm
Millimole of CH3COONa = 2.5
After nutrilisation,
Concentration of CH3COOH = $[C{H_3}COOH] = {{5 - 2.5} \over {75}} = {1 \over {30}}$
Concentration of CH3COONa = $[C{H_3}COONa] = {{ 2.5} \over {75}} = {1 \over {30}}$
${P^H} = {P^{Ka}} + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$
$ = 4.76 + \log {{{1 \over {30}}} \over {{1 \over {30}}}}$
$ = 4.76 + \log (1)$
$ = 4.76 + 0$
$ = 4.76$
$ = 4.76 \times {10^{ - 2}}$
2022
JEE Advanced
Numerical
JEE Advanced 2022 Paper 2 Online
Concentration of $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{Na}_{2} \mathrm{SO}_{4}$ in a solution is $1 \mathrm{M}$ and $1.8 \times 10^{-2} \mathrm{M}$, respectively. Molar solubility of $\mathrm{PbSO}_{4}$ in the same solution is $\mathrm{X} \times 10^{-\mathrm{Y}} \mathrm{M}$ (expressed in scientific notation). The value of $Y$ is ________.
[Given: Solubility product of $\mathrm{PbSO}_{4}\left(K_{s p}\right)=1.6 \times 10^{-8}$. For $\mathrm{H}_{2} \mathrm{SO}_{4}, K_{a l}$ is very large and $\left.K_{a 2}=1.2 \times 10^{-2}\right]$
Correct Answer: 6
Explanation:
Given,
Conentration of $\mathrm{H}_2 \mathrm{SO}_4=1 \mathrm{M}$
Conentration of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ = $1.8 \times 10^{-2} \mathrm{M}$
From the above two equations, we get
$ \left[\mathrm{H}^{+}\right]=1 \mathrm{M} \text { and }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2} $
$ K_c=\frac{1.8 \times 10^{-2} \times 1}{1}=1.8 \times 10^{-2} $
and it is given that $\mathrm{K}_{a_2}\left(\mathrm{Q}_c\right)=1.2 \times 10^{-2} \mathrm{M}$
Since, $\mathrm{K}_{a_2}$ (i.e., $\mathrm{Q}_c$ ) $<\mathrm{K}_c$
$\therefore$ Rather than dissociation of $\mathrm{HSO}_4^{-}$into $\mathrm{H}^{+}$and $\mathrm{SO}_4^{2-}$ ions, association between already present $\mathrm{H}^{+}$and $\mathrm{SO}_4^{2-}$ will take place.
Assuming ' $x$ ' mol/L of $\mathrm{SO}_4^{2-}$ and $\mathrm{H}^{+}$combines to form $\mathrm{HSO}_4^{-}$
$\begin{aligned} & \mathrm{K}_{a 2}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\left[\mathrm{HSO}_4^{-}\right]} \\\\ & 1.2 \times 10^{-2}=\frac{(1-x)\left(1.8 \times 10^{-2}-x\right)}{(1+x)} \\\\ & \because x<<1, \text { so }(1+x) \approx 1 \text { and }(1-x) \approx 1 \\\\ & 1.2 \times 10^{-2}=1.8 \times 10^{-2}-x \\\\ & x=\left(1.8 \times 10^{-2}\right)-\left(1.2 \times 10^{-2}\right) \\\\ & x=0.6 \times 10^{-2} \mathrm{M} \\\\ & \text { So, }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2}-x \\\\ & {\left[\mathrm{SO}_4^{2-}\right] }=\left(1.8 \times 10^{-2}\right)-\left(0.6 \times 10^{-2}\right) \\\\ & {\left[\mathrm{SO}_4^{2-}\right] }=1.2 \times 10^{-2} \mathrm{M}\end{aligned}$
Given, $\quad \mathrm{K}_{s p}=1.6 \times 10^{-8}$
$ \therefore $ $y\left(1.2 \times 10^{-2}+y\right)=1.6 \times 10^{-8}$
Since, $y<<1$, So $1.2 \times 10^{-2}+y \approx 1.2 \times 10^{-2}$
So, $y \times 1.2 \times 10^{-2}=1.6 \times 10^{-8}$
$ \Rightarrow $ $y =\frac{1.6 \times 10^{-8}}{1.2 \times 10^{-2}}$
$ \Rightarrow $ $ y =1.33 \times 10^{-6} $
$ X \times 10^{-Y} \mathrm{M} =1.33 \times 10^{-6} \mathrm{M} $
So, $\mathrm{Y}=6$
Hence, the value of $Y$ is 6 .
Conentration of $\mathrm{H}_2 \mathrm{SO}_4=1 \mathrm{M}$
Conentration of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ = $1.8 \times 10^{-2} \mathrm{M}$
From the above two equations, we get
$ \left[\mathrm{H}^{+}\right]=1 \mathrm{M} \text { and }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2} $
$ K_c=\frac{1.8 \times 10^{-2} \times 1}{1}=1.8 \times 10^{-2} $
and it is given that $\mathrm{K}_{a_2}\left(\mathrm{Q}_c\right)=1.2 \times 10^{-2} \mathrm{M}$
Since, $\mathrm{K}_{a_2}$ (i.e., $\mathrm{Q}_c$ ) $<\mathrm{K}_c$
$\therefore$ Rather than dissociation of $\mathrm{HSO}_4^{-}$into $\mathrm{H}^{+}$and $\mathrm{SO}_4^{2-}$ ions, association between already present $\mathrm{H}^{+}$and $\mathrm{SO}_4^{2-}$ will take place.
Assuming ' $x$ ' mol/L of $\mathrm{SO}_4^{2-}$ and $\mathrm{H}^{+}$combines to form $\mathrm{HSO}_4^{-}$
$\begin{aligned} & \mathrm{K}_{a 2}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\left[\mathrm{HSO}_4^{-}\right]} \\\\ & 1.2 \times 10^{-2}=\frac{(1-x)\left(1.8 \times 10^{-2}-x\right)}{(1+x)} \\\\ & \because x<<1, \text { so }(1+x) \approx 1 \text { and }(1-x) \approx 1 \\\\ & 1.2 \times 10^{-2}=1.8 \times 10^{-2}-x \\\\ & x=\left(1.8 \times 10^{-2}\right)-\left(1.2 \times 10^{-2}\right) \\\\ & x=0.6 \times 10^{-2} \mathrm{M} \\\\ & \text { So, }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2}-x \\\\ & {\left[\mathrm{SO}_4^{2-}\right] }=\left(1.8 \times 10^{-2}\right)-\left(0.6 \times 10^{-2}\right) \\\\ & {\left[\mathrm{SO}_4^{2-}\right] }=1.2 \times 10^{-2} \mathrm{M}\end{aligned}$
Given, $\quad \mathrm{K}_{s p}=1.6 \times 10^{-8}$
$ \therefore $ $y\left(1.2 \times 10^{-2}+y\right)=1.6 \times 10^{-8}$
Since, $y<<1$, So $1.2 \times 10^{-2}+y \approx 1.2 \times 10^{-2}$
So, $y \times 1.2 \times 10^{-2}=1.6 \times 10^{-8}$
$ \Rightarrow $ $y =\frac{1.6 \times 10^{-8}}{1.2 \times 10^{-2}}$
$ \Rightarrow $ $ y =1.33 \times 10^{-6} $
$ X \times 10^{-Y} \mathrm{M} =1.33 \times 10^{-6} \mathrm{M} $
So, $\mathrm{Y}=6$
Hence, the value of $Y$ is 6 .
2022
JEE Advanced
Numerical
JEE Advanced 2022 Paper 1 Online
A solution is prepared by mixing $0.01 \mathrm{~mol}$ each of $\mathrm{H}_{2} \mathrm{CO}_{3}, \mathrm{NaHCO}_{3}, \mathrm{Na}_{2} \mathrm{CO}_{3}$, and $\mathrm{NaOH}$ in $100 \mathrm{~mL}$ of water. $p \mathrm{H}$ of the resulting solution is _________.
[Given: $p \mathrm{~K}_{\mathrm{a} 1}$ and $p \mathrm{~K}_{\mathrm{a} 2}$ of $\mathrm{H}_{2} \mathrm{CO}_{3}$ are $6.37$ and 10.32, respectively; $\log 2=0.30$ ]
Correct Answer: 10.00TO10.04
Explanation:
First acid base reaction between H2CO3 and NaOH takes place.
In the final solution, we have 0.01 mole Na2CO3 and 0.02 moles of NaHCO3.
Here, we have a buffer of NaHCO3 and Na2CO3.
$\therefore$ $pH = p{K_{{a_2}}} + \log {{[Salt]} \over {[Acid]}}$
$ = 10.32 + \log {{\left( {{{0.01} \over {0.1}}} \right)} \over {\left( {{{0.02} \over {0.1}}} \right)}}$
$ = 10.32 + \log {1 \over 2}$
$ = 10.32 - \log 2$
$ = 10.32 - 0.3$
$ = 10.02$
$\therefore$ $pH = 10.02$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
Given below are two statements.
Statement I : In the titration between strong acid and weak base methyl orange is suitable as an indicator.
Statement II : For titration of acetic acid with NaOH phenolphthalein is not a suitable indicator.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : In the titration between strong acid and weak base methyl orange is suitable as an indicator.
Statement II : For titration of acetic acid with NaOH phenolphthalein is not a suitable indicator.
In the light of the above statements, choose the most appropriate answer from the options given below :
A.
Statement I is false but Statement II is true
B.
Statement I is true but Statement II is false
C.
Both Statement I and Statement II are true
D.
Both Statement I and Statement II are false
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Evening Shift
A solution is 0.1 M in Cl$-$ and 0.001 M in CrO$_4^{2 - }$. Solid AgNO3 is gradually added to it. Assuming that the addition does not change in volume and Ksp(AgCl) = 1.7 $\times$ 10$-$10 M2 and Ksp(Ag2CrO4) = 1.9 $\times$ 10$-$12 M3.
Select correct statement from the following :
Select correct statement from the following :
A.
AgCl precipitates first because its Ksp is high.
B.
Ag2CrO4 precipitates first as its Ksp is low.
C.
Ag2CrO4 precipitates first because the amount of Ag+ needed is low.
D.
AgCl will precipitate first as the amount of Ag+ needed to precipitate is low.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
Given below are two statements : One is labelled as Assertion A and the other labelled as reason R
Assertion A : During the boiling of water having temporary hardness, Mg(HCO3)2 is converted to MgCO3.
Reason R : The solubility product of Mg(OH)2 is greater than that of MgCO3.
In the light of the above statements, choose the most appropriate answer from the options given below :
Assertion A : During the boiling of water having temporary hardness, Mg(HCO3)2 is converted to MgCO3.
Reason R : The solubility product of Mg(OH)2 is greater than that of MgCO3.
In the light of the above statements, choose the most appropriate answer from the options given below :
A.
Both A and R are true but R is not the correct explanation of A
B.
A is true but R is false
C.
Both A and R are true and R is the correct explanation of A
D.
A is false but R is true
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
Which of the following compound CANNOT act as a Lewis base?
A.
SF4
B.
NF3
C.
ClF3
D.
PCl5
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
The solubility of Ca(OH)2 in water is :
[Given : The solubility product of Ca(OH)2 in water = 5.5 $\times$ 10$-$6]
[Given : The solubility product of Ca(OH)2 in water = 5.5 $\times$ 10$-$6]
A.
1.11 $\times$ 10$-$6
B.
1.11 $\times$ 10$-$2
C.
1.77 $\times$ 10$-$6
D.
1.77 $\times$ 10$-$2
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is : [Assume : No cyano complex is formed; Ksp(AgCN) = 2.2 $\times$ 10$-$16 and Ka(HCN) = 6.2 $\times$ 10$-$10]
A.
1.9 $\times$ 10$-$5
B.
1.6 $\times$ 10$-$6
C.
2.2 $\times$ 10$-$16
D.
0.625 $\times$ 10$-$6
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 1st September Evening Shift
The molar solubility of Zn(OH)2 in 0.1 M NaOH solution is x $\times$ 10$-$18 M. The value of x is _________ (Nearest integer)
(Given : The solubility product of Zn(OH)2 is 2 $\times$ 10$-$20)
(Given : The solubility product of Zn(OH)2 is 2 $\times$ 10$-$20)
Correct Answer: 2
Explanation:
Due to common-ion effect (presence of NaOH) the concentration of OH$-$ will be (2S + 0.1) $\approx$ 0.1
($\because$ 0.1 > > 2 S)
$\therefore$ Solubility of product,
${K_{sp}} = {(0.1)^2} \times S$
$2 \times {10^{ - 20}} = 0.01 \times S$
$ \Rightarrow S = {{2 \times {{10}^{ - 20}}} \over {0.01}} = 2 \times {10^{ - 18}}$
$\therefore$ x = 2
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M NaOH is x $\times$ 10$-$4. The value of x is ____________. (Nearest integer) [log 2.5 = 0.3979]
Correct Answer: 6021
Explanation:
$[HCl] = {{20} \over {80}} = {1 \over 4}M = 2.5 \times {10^{ - 1}}M$
pH = $-$log 2.15 $\times$ 10-1 = 1 $-$ 0.3979 = 0.6021
pH = 6021 $\times$ 10-4
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
A3B2 is a sparingly soluble salt of molar mass M (g mol$-$1) and solubility x g L$-$1. The solubility product satisfies ${K_{sp}} = a{\left( {{x \over M}} \right)^5}$. The value of a is _____________. (Integer answer)
Correct Answer: 108
Explanation:
KSP = (3s)3 (2s)2
KSP = 108 s5 & s = (x/M)
KSP = 108${\left( {{x \over M}} \right)^5}$
given ${K_{sp}} = a{\left( {{x \over M}} \right)^5}$
comparing a = 108
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
The solubility of CdSO4 in water is 8.0 $\times$ 10$-$4 mol L$-$1. Its solubility in 0.01 M H2SO4 solution is __________ $\times$ 10$-$6 mol L$-$1. (Round off to the Nearest Integer). (Assume that solubility is much less than 0.01 M)
Correct Answer: 64
Explanation:
In pure water
$ \therefore $ Ksp = (s)2 = (8 $\times$ 10-4)2 = 64 $\times$ 10-8
In H2SO4 solution,
As S1 < < 0.01 so, S1 + 0.01 $ \simeq $ 0.01
$ \therefore $ Ksp = [Cd+2] [So$_4^{ - 2}$]
$ \Rightarrow $ 64 $\times$ 10-8 = S1 $\times$ 0.01
$ \Rightarrow $ S1 = 64 $\times$ 10-6
$ \therefore $ Ksp = (s)2 = (8 $\times$ 10-4)2 = 64 $\times$ 10-8
In H2SO4 solution,
As S1 < < 0.01 so, S1 + 0.01 $ \simeq $ 0.01
$ \therefore $ Ksp = [Cd+2] [So$_4^{ - 2}$]
$ \Rightarrow $ 64 $\times$ 10-8 = S1 $\times$ 0.01
$ \Rightarrow $ S1 = 64 $\times$ 10-6
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Morning Shift
In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is ___________ M. (Round off to the Nearest Integer). [Given : pKa (acetic acid) = 4.74]
Correct Answer: 10
Explanation:
$C{H_3}COOH + C{H_3}COONa$ [Acidic Buffer]
${pH} = {p{ka}} + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$
$ \Rightarrow 5.74 = 4.74 + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$
$ \Rightarrow 1 = \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$
$ \Rightarrow {{[C{H_3}COONa]} \over {[C{H_3}COOH]}} = 10$
$ \Rightarrow [C{H_3}COONa] = 10 \times 1 = 10$
${pH} = {p{ka}} + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$
$ \Rightarrow 5.74 = 4.74 + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$
$ \Rightarrow 1 = \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$
$ \Rightarrow {{[C{H_3}COONa]} \over {[C{H_3}COOH]}} = 10$
$ \Rightarrow [C{H_3}COONa] = 10 \times 1 = 10$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
Sulphurous acid (H2SO3) has Ka1 = 1.7 $\times$ 10$-$2 and Ka2 = 6.4 $\times$ 10$-$8. The pH of 0.588 M H2SO3 is __________. (Round off to the Nearest Integer).
Correct Answer: 1
Explanation:
$K{a_1}$ of ${H_2}S{O_3} > > K{a_2}$ of ${H_2}S{O_3}$
$ \therefore $ The contribution of H+ from 2nd dissociation of H2SO3 can be neglected.
$ \Rightarrow $ ${{c{\alpha ^2}} \over {1 - \alpha }} = 1.7 \times {10^{ - 2}}$
$ \Rightarrow {{0.588{\alpha ^2}} \over {1 - \alpha }} = 1.7 \times {10^{ - 2}}$
$ \Rightarrow 58.8{\alpha ^2} = 1.7 - 1.7\alpha $
$ \Rightarrow 58.8{\alpha ^2} + 1.7\alpha - 1.7 = 0$
$\alpha = {{ - 1.7 + \sqrt {{{1.7}^2} + 4 \times 1.7 \times 58.8} } \over {2 \times 58.8}} = 0.156$
$[{H^ + }] = c\alpha = 0.092$
$pH = - \log [{H^ + }]$
$ = 1.036$
$ \approx 1$
$ \therefore $ The contribution of H+ from 2nd dissociation of H2SO3 can be neglected.
$ \Rightarrow $ ${{c{\alpha ^2}} \over {1 - \alpha }} = 1.7 \times {10^{ - 2}}$
$ \Rightarrow {{0.588{\alpha ^2}} \over {1 - \alpha }} = 1.7 \times {10^{ - 2}}$
$ \Rightarrow 58.8{\alpha ^2} = 1.7 - 1.7\alpha $
$ \Rightarrow 58.8{\alpha ^2} + 1.7\alpha - 1.7 = 0$
$\alpha = {{ - 1.7 + \sqrt {{{1.7}^2} + 4 \times 1.7 \times 58.8} } \over {2 \times 58.8}} = 0.156$
$[{H^ + }] = c\alpha = 0.092$
$pH = - \log [{H^ + }]$
$ = 1.036$
$ \approx 1$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
Two salts A2X and MX have the same value of solubility product of 4.0 $\times$ 10$-$12. The ratio of their molar solubilities i.e. ${{S({A_2}X)} \over {S(MX)}}$ = __________. (Round off to the Nearest Integer)
Correct Answer: 50
Explanation:
For A2X
$\matrix{ {{A_2}X} & \to & {2{A^ + }} & {{X^{2 - }}} \cr {} & {} & {2{S_1}} & {{S_1}} \cr } $
${K_{sp}} = 4S_1^3 = 4 \times {10^{ - 12}}$
S1 = 10$-$4
for MX
$\matrix{ {MX} & \to & {{M^ + }} & {{X^ - }} \cr {} & {} & {{S_2}} & {{S_2}} \cr } $
${K_{sp}} = S_2^2 = 4 \times {10^{ - 12}}$
S2 = 2 $\times$ 10$-$6
so, ${{{S_{{A_2}X}}} \over {{S_{MX}}}} = {{{{10}^{ - 4}}} \over {2 \times {{10}^{ - 6}}}} = 50$
$\matrix{ {{A_2}X} & \to & {2{A^ + }} & {{X^{2 - }}} \cr {} & {} & {2{S_1}} & {{S_1}} \cr } $
${K_{sp}} = 4S_1^3 = 4 \times {10^{ - 12}}$
S1 = 10$-$4
for MX
$\matrix{ {MX} & \to & {{M^ + }} & {{X^ - }} \cr {} & {} & {{S_2}} & {{S_2}} \cr } $
${K_{sp}} = S_2^2 = 4 \times {10^{ - 12}}$
S2 = 2 $\times$ 10$-$6
so, ${{{S_{{A_2}X}}} \over {{S_{MX}}}} = {{{{10}^{ - 4}}} \over {2 \times {{10}^{ - 6}}}} = 50$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
The pH of ammonium phosphate solution, if pka of phosphoric acid and pkb of ammonium hydroxide are 5.23 and 4.75 respectively, is ___________.
Correct Answer: 7
Explanation:
Since (NH4)3PO4 is salt of weak acid (H3PO4) & weak base (NH4OH).
pH = 7 + ${1 \over 2}$(pka $-$ pkb)
= 7 + ${1 \over 2}$ (5.23 $-$ 4.75)
= 7.24 $ \approx $ 7.
pH = 7 + ${1 \over 2}$(pka $-$ pkb)
= 7 + ${1 \over 2}$ (5.23 $-$ 4.75)
= 7.24 $ \approx $ 7.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
The solubility product of PbI2 is 8.0 $\times$ 10$-$9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x $\times$ 10$-$6. mol/L. The value of x is __________. (Rounded off to the nearest integer) [Given $\sqrt 2 $ = 1.41]
Correct Answer: 141
Explanation:
Given, ${[{K_{sp}}]_{Pb{l_2}}} = 8 \times {10^{ - 9}}$
To calculate solubility of Pbl2 in 0.1 M solution of Pb(NO3)2,
(I) $\mathop {Pb{{(N{O_3})}_2}}\limits_{0. 1 M} \to \mathop {P{b^{2 + }}(aq)}\limits_{0.1 M} + \mathop {2NO_3^ - (aq)}\limits_{0. 2 M} $
(II) $Pb{I_2}(s)$ $\rightleftharpoons$ $\mathop {P{b^{2 + }}(aq)}\limits_S + \mathop {2{I^ - }(aq)}\limits_{2S} $
$\therefore$ [Pb2+] = S + 0.1 $\approx$ 0.1
$\because$ S < < 0.1
Now, Ksp = 8 $\times$ 10$-$9
[Pb2] [I$-$]2 = 8 $\times$ 10$-$9
0.1 $\times$ (2S)2 = 8 $\times$ 10$-$9
4S2 = 8 $\times$ 10$-$8 $\Rightarrow$ S = 141 $\times$ 10$-$6 M
$ \therefore $ x = 141
To calculate solubility of Pbl2 in 0.1 M solution of Pb(NO3)2,
(I) $\mathop {Pb{{(N{O_3})}_2}}\limits_{0. 1 M} \to \mathop {P{b^{2 + }}(aq)}\limits_{0.1 M} + \mathop {2NO_3^ - (aq)}\limits_{0. 2 M} $
(II) $Pb{I_2}(s)$ $\rightleftharpoons$ $\mathop {P{b^{2 + }}(aq)}\limits_S + \mathop {2{I^ - }(aq)}\limits_{2S} $
$\therefore$ [Pb2+] = S + 0.1 $\approx$ 0.1
$\because$ S < < 0.1
Now, Ksp = 8 $\times$ 10$-$9
[Pb2] [I$-$]2 = 8 $\times$ 10$-$9
0.1 $\times$ (2S)2 = 8 $\times$ 10$-$9
4S2 = 8 $\times$ 10$-$8 $\Rightarrow$ S = 141 $\times$ 10$-$6 M
$ \therefore $ x = 141
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
Arrange the following solutions in the
decreasing order of pOH :
(A) 0.01 M HCl
(B) 0.01 M NaOH
(C) 0.01 M CH3COONa
(D) 0.01 M NaCl
(A) 0.01 M HCl
(B) 0.01 M NaOH
(C) 0.01 M CH3COONa
(D) 0.01 M NaCl
A.
(B) > (C) > (D) > (A)
B.
(A) > (D) > (C) > (B)
C.
(A) > (C) > (D) > (B)
D.
(B) > (D) > (C) > (A)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
100 mL of 0.1 M HCl is taken in a beaker and
to it 100 mL of 0.1 M NaOH is added in steps
of 2 mL and the pH is continuously measured.
Which of the following graphs correctly depicts
the change in pH?
A.
B.
C.
D.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
An acidic buffer is obtained on mixing :
A.
100 mL of 0.1 M HCl and 200 mL of 0.1 M CH3COONa
B.
100 mL of 0.1 M HCl and 200 mL of 0.1 M NaCl
C.
100 mL of 0.1 M CH3COOH and 100 mL of 0.1 M NaOH
D.
100 mL of 0.1 M CH3COOH and 200 mL of 0.1 M NaOH
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
For the following Assertion and Reason, the
correct option is
Assertion (A): When Cu (II) and sulphide ions are mixed, they react together extremely quickly to give a solid.
Reason (R): The equilibrium constant of
Cu2+(aq) + S2–(aq) ⇌ CuS(s) is high because the solubility product is low.
Assertion (A): When Cu (II) and sulphide ions are mixed, they react together extremely quickly to give a solid.
Reason (R): The equilibrium constant of
Cu2+(aq) + S2–(aq) ⇌ CuS(s) is high because the solubility product is low.
A.
(A) is false and (R) is true.
B.
Both (A) and (R) are true but (R) is not the
explanation for (A).
C.
Both (A) and (R) are true and (R) is the
explanation for (A).
D.
Both (A) and (R) are false.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
The solubility product of Cr(OH)3 at 298 K is
6.0 × 10–31. The concentration of hydroxide ions
in a saturated solution of Cr(OH)3 will be :
A.
(2.22 × 10–31)1/4
B.
(4.86 × 10–29)1/4
C.
(18 × 10–31)1/4
D.
(18 × 10–31)1/2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
The Ksp for the following dissociation is
1.6 × 10–5
$PbC{l_{2(s)}} \leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ - $
Which of the following choices is correct for a mixture of 300 mL 0.134 M Pb(NO3)2 and 100 mL 0.4 M NaCl ?
$PbC{l_{2(s)}} \leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ - $
Which of the following choices is correct for a mixture of 300 mL 0.134 M Pb(NO3)2 and 100 mL 0.4 M NaCl ?
A.
Q > Ksp
B.
Not enough data provided
C.
Q < Ksp
D.
Q = Ksp
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
For the following Assertion and Reason, the
correct option is :
Assertion : The pH of water increases with increase in temperature.
Reason : The dissociation of water into H+ and OH– is an exothermic reaction.
Assertion : The pH of water increases with increase in temperature.
Reason : The dissociation of water into H+ and OH– is an exothermic reaction.
A.
Both assertion and reason are false.
B.
Both assertion and reason are true, but the
reason is not the correct explanation for the
assertion.
C.
Both assertion and reason are true, and the
reason is the correct explanation for the
assertion.
D.
Assertion is not true, but reason is true.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
The strength of an aqueous NaOH solution is most accurately determined by titrating :
(Note : consider that an appropriate indicator is used)
(Note : consider that an appropriate indicator is used)
A.
Aq. NaOH in a pipette and aqueous oxalic acid in a burette
B.
Aq. NaOH in a burette and aqueous oxalic acid in a conical flask
C.
Aq. NaOH in a volumetric flask and concentrated H2SO4 in a conical flask
D.
Aq. NaOH in a burette and concentrated H2SO4 in a conical flask
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively :
A.
XY, 2 × 10–6 M3
B.
XY2, 1 × 10–9 M3
C.
XY2, 4 × 10–9 M3
D.
X2Y, 2 × 10–9 M3
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Evening Slot
If the solubility product of AB2 is 3.20 $ \times $ 10–11 M3,
then the solubility of AB2 in pure water is
_____ $ \times $ 10–4 mol L–1.
[Assuming that neither kind of ion reacts with water]
[Assuming that neither kind of ion reacts with water]
Correct Answer: 2
Explanation:
| AB2 | ⇌ | A2+(aq) | + | 2B-(aq) |
|---|---|---|---|---|
| s | 2s |
Ksp = 4s3 = 3.2 × 10–11
$ \Rightarrow $ s3 = 8 × 10–12
$ \Rightarrow $ s = 2 × 10–4
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 5th September Morning Slot
A soft drink was bottled with a partial pressure of CO2
of 3 bar over the liquid at room temperature.
The partial pressure of CO2
over the solution approaches a value of 30 bar when 44 g of CO2
is
dissolved in 1 kg of water at room temperature. The approximate pH of the soft drink is ______ $ \times $ 10–1.
(First dissociation constant of
H2CO3 = 4.0 $ \times $ 10–7; log 2 = 0.3; density
of the soft drink = 1 g mL–1) .
(First dissociation constant of
H2CO3 = 4.0 $ \times $ 10–7; log 2 = 0.3; density
of the soft drink = 1 g mL–1) .
Correct Answer: 37
Explanation:
CO2
+ H2O $ \to $ H2CO3
At 30 bar pressure mass of CO2 in 1 kg water = 44 gm
At 3 bar pressure mass of CO2 in 1 kg water = 4.4 gm
$ \therefore $ Moles of CO2 in 1 kg water = ${{4.4} \over {44}}$ = 0.1
4.0 $ \times $ 10–7 = ${{0.1{\alpha ^2}} \over {1 - \alpha }}$
${1 - \alpha }$ $ \simeq $ 1
$ \Rightarrow $ 0.1${{\alpha ^2}}$ = 4 $ \times $ 10-7
$ \Rightarrow $ $\alpha $ = 2 $ \times $ 10-3
[H+] = 0.1$\alpha $ = 2 $ \times $ 10-4
$ \therefore $ pH = –[– 4 × log(2)] = 3.7 = 37 × 10–1
At 30 bar pressure mass of CO2 in 1 kg water = 44 gm
At 3 bar pressure mass of CO2 in 1 kg water = 4.4 gm
$ \therefore $ Moles of CO2 in 1 kg water = ${{4.4} \over {44}}$ = 0.1
| H2CO3 | ⇌ | H+ | + | HCO3- | |
|---|---|---|---|---|---|
| t = 0 | 0.1 | 0 | 0 | ||
| t = teq | 0.1(1 - $\alpha $) | 0.1$\alpha $ | 0.1$\alpha $ |
4.0 $ \times $ 10–7 = ${{0.1{\alpha ^2}} \over {1 - \alpha }}$
${1 - \alpha }$ $ \simeq $ 1
$ \Rightarrow $ 0.1${{\alpha ^2}}$ = 4 $ \times $ 10-7
$ \Rightarrow $ $\alpha $ = 2 $ \times $ 10-3
[H+] = 0.1$\alpha $ = 2 $ \times $ 10-4
$ \therefore $ pH = –[– 4 × log(2)] = 3.7 = 37 × 10–1
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Evening Slot
3 g of acetic acid is added to 250 mL of 0.1 M HCL and the solution made up to 500 mL. To 20 mL
of this solutions ${1 \over 2}$ mL of 5 M NaOH is added. The pH of the solution is __________.
[Given : pKa of acetic acid = 4.75, molar mass of acetic of acid = 60 g/mol, log 3 = 0.4771] Neglect any changes in volume.
[Given : pKa of acetic acid = 4.75, molar mass of acetic of acid = 60 g/mol, log 3 = 0.4771] Neglect any changes in volume.
Correct Answer: 5.22TO5.24
Explanation:
milimole of acetic acid in 20 ml
= ${3 \over {60}} \times 1000 \times {{20} \over {500}}$ = 2
milimole of HCl in 20 ml = 25 $ \times $ ${{20} \over {500}}$ = 1
milimole of NaOH 20 ml = ${1 \over 2} \times 5$ = 2.5
pH = pKa + log${{1.5} \over {0.5}}$
= 4.74 + log 3 = 5.22
= ${3 \over {60}} \times 1000 \times {{20} \over {500}}$ = 2
milimole of HCl in 20 ml = 25 $ \times $ ${{20} \over {500}}$ = 1
milimole of NaOH 20 ml = ${1 \over 2} \times 5$ = 2.5
| NaOH | + | CH3COOH | $ \to $ | CH3COONa | + | H2O |
|---|---|---|---|---|---|---|
| 1.5 | 2 | 0 | 0 | |||
| 0 | 0.5 | 1.5 |
pH = pKa + log${{1.5} \over {0.5}}$
= 4.74 + log 3 = 5.22
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Morning Slot
Two solutions, A and B, each of 100L was made by dissolving 4g of NaOH and 9.8 g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40L of solution A and 10L of solution B is :
Correct Answer: 10.6
Explanation:
In 100 L solution H2SO4 present = 9.8 gm
$ \therefore $ In 10 L solution H2SO4 present = $9.8 \times {{10} \over {100}}$ gm
$ \Rightarrow $ In 10 L solution moles of H2SO4 present = ${{9.8} \over {98}} \times {{10} \over {100}}$
In one molecule of H2SO4 two H+ ion present.
$ \therefore $ In 10 L solution moles of H+ present = 2$ \times $${{9.8} \over {98}} \times {{10} \over {100}}$ = 0.02 moles
Also In 100 L solution NaOH present = 4 gm
$ \therefore $ In 40 L solution NaOH present = $4 \times {{40} \over {100}}$
$ \Rightarrow $ In 40 L solution moles of NaOH present = ${4 \over {40}} \times {{40} \over {100}}$
In one molecule of NaOH one OH- ion present.
$ \therefore $ In 40 L solution moles of OH- ion present = ${4 \over {40}} \times {{40} \over {100}}$ = 0.04 moles
As moles of OH- ion is more than H+ ion, so solution is basic.
$ \therefore $ Final Conc. of OH– = ${{0.04 - 0.02} \over {40 + 10}}$ = 4 $ \times $ 10-4
$ \therefore $ pOH = – log (4 ×10–4) = 3.4
pH = 14 – 3.4 = 10.6
$ \therefore $ In 10 L solution H2SO4 present = $9.8 \times {{10} \over {100}}$ gm
$ \Rightarrow $ In 10 L solution moles of H2SO4 present = ${{9.8} \over {98}} \times {{10} \over {100}}$
In one molecule of H2SO4 two H+ ion present.
$ \therefore $ In 10 L solution moles of H+ present = 2$ \times $${{9.8} \over {98}} \times {{10} \over {100}}$ = 0.02 moles
Also In 100 L solution NaOH present = 4 gm
$ \therefore $ In 40 L solution NaOH present = $4 \times {{40} \over {100}}$
$ \Rightarrow $ In 40 L solution moles of NaOH present = ${4 \over {40}} \times {{40} \over {100}}$
In one molecule of NaOH one OH- ion present.
$ \therefore $ In 40 L solution moles of OH- ion present = ${4 \over {40}} \times {{40} \over {100}}$ = 0.04 moles
As moles of OH- ion is more than H+ ion, so solution is basic.
$ \therefore $ Final Conc. of OH– = ${{0.04 - 0.02} \over {40 + 10}}$ = 4 $ \times $ 10-4
$ \therefore $ pOH = – log (4 ×10–4) = 3.4
pH = 14 – 3.4 = 10.6
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
A solution of 0.1 M weak base (B) is titrated with 0.1 M of a strong acid (HA). The variation of pH of the solution with the volume of HA added is shown in the figure below. What is the pKb of the base? The neutralisation reaction is given by
$B + HA\buildrel {} \over \longrightarrow B{H^ + } + {A^ - }$
$B + HA\buildrel {} \over \longrightarrow B{H^ + } + {A^ - }$
Correct Answer: 3
Explanation:
From the given diagram, 63 mL volume of HA used till equivalence point. At half of equivalence point, solution will be basic buffer with B and BH+.
$pOH = p{K_b} + \log {{[B{H^ + }]} \over {[B]}}$
At half equivalence point :
[BH+] = [B] ($ \because $ pH = 11)
Therefore, pOH = pKb = 14 $-$ 11 = 3
$ \because $ pKb = 3.00
$pOH = p{K_b} + \log {{[B{H^ + }]} \over {[B]}}$
At half equivalence point :
[BH+] = [B] ($ \because $ pH = 11)
Therefore, pOH = pKb = 14 $-$ 11 = 3
$ \because $ pKb = 3.00
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
An acidified solution of 0.05 M Zn2+ is saturated with 0.1 M H2S. What is the minimum molar concentration (M) of H+ required to prevent the precipitation of ZnS?
Use Ksp(ZnS) = 1.25 $ \times $ 10$-$22 and overall dissociation constant of
H2S, Knet = K1K2 = 1 $ \times $ 10-21.
Use Ksp(ZnS) = 1.25 $ \times $ 10$-$22 and overall dissociation constant of
H2S, Knet = K1K2 = 1 $ \times $ 10-21.
Correct Answer: 0.2
Explanation:
$ZnS(s)\buildrel {} \over
\longrightarrow \mathop {Z{n^{2 + }}(aq)}\limits_{(0.05\,M)} + {S^{2 - }}(aq)$
${K_{sp}}(ZnS) = [Z{n^{2 + }}][{S^{2 - }}] = 1.25 \times {10^{ - 22}}$
$0.05 \times [{S^{2 - }}] = 1.25 \times {10^{ - 22}}$
$ \Rightarrow [{S^{2 - }}] = {{1.25 \times {{10}^{ - 22}}} \over {0.05}} \Rightarrow 25 \times {10^{ - 22}}M$
For H2S, $\mathop {{H_2}S}\limits_{(0.1M)} \buildrel {} \over \longrightarrow 2{H^ + } + \mathop {{S^{ - 2}}}\limits_{(25 \times {{10}^{ - 22}}M)} $
${K_{net}} = 1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2}[{S^{2 - }}]} \over {[{H_2}S]}}$
$1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2} \times 25 \times {{10}^{ - 22}}} \over {[0.1]}}$
${[{H^ + }]^2} = {1 \over {25}}$
$[{H^ + }] = {1 \over 5} = 0.2M$
${K_{sp}}(ZnS) = [Z{n^{2 + }}][{S^{2 - }}] = 1.25 \times {10^{ - 22}}$
$0.05 \times [{S^{2 - }}] = 1.25 \times {10^{ - 22}}$
$ \Rightarrow [{S^{2 - }}] = {{1.25 \times {{10}^{ - 22}}} \over {0.05}} \Rightarrow 25 \times {10^{ - 22}}M$
For H2S, $\mathop {{H_2}S}\limits_{(0.1M)} \buildrel {} \over \longrightarrow 2{H^ + } + \mathop {{S^{ - 2}}}\limits_{(25 \times {{10}^{ - 22}}M)} $
${K_{net}} = 1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2}[{S^{2 - }}]} \over {[{H_2}S]}}$
$1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2} \times 25 \times {{10}^{ - 22}}} \over {[0.1]}}$
${[{H^ + }]^2} = {1 \over {25}}$
$[{H^ + }] = {1 \over 5} = 0.2M$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
The decreasing order of electrical conductivity of the following aqueous solutions is :
0.1 M Formic acid (A),
0.1 M Acetic acid (B),
0.1 M Benzoic acid (C)
0.1 M Formic acid (A),
0.1 M Acetic acid (B),
0.1 M Benzoic acid (C)
A.
A > B > C
B.
C > B > A
C.
A > C > B
D.
C > A > B
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
The molar solubility of Cd(OH)2 is 1.84 × 10–5
M in water. The expected solubility of Cd(OH)2 in a buffer
solution of pH = 12 is :
A.
2.49 × 10–10 M
B.
1.84 × 10–9
M
C.
6.23 × 10–11 M
D.
${{2.49} \over {1.84}} \times {10^{ - 9}}M$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
What is the molar solubility of Al(OH)3 in 0.2 M NaOH solution ? Given that, solubility product of Al(OH)3 = 2.4 × 10–24
:
A.
3 × 10–22
B.
3 × 10–19
C.
12 × 10–21
D.
12 × 10–22
