Ionic Equilibrium
Millimoles of calcium hydroxide required to produce 100 mL of the aqueous solution of pH 12 is $x\times10^{-1}$. The value of $x$ is ___________ (Nearest integer).
Assume complete dissociation.
Explanation:
Moles of $\mathrm{Ca}(\mathrm{OH})_2$ in $1000 \mathrm{~mL}=5 \times 10^{-3}$
Millimoles in $100 \mathrm{~mL}=5 \times 10^{-1}$
A litre of buffer solution contains 0.1 mole of each of NH$_3$ and NH$_4$Cl. On the addition of 0.02 mole of HCl by dissolving gaseous HCl, the pH of the solution is found to be _____________ $\times$ 10$^{-3}$ (Nearest integer)
[Given : $\mathrm{pK_b(NH_3)=4.745}$
$\mathrm{\log2=0.301}$
$\mathrm{\log3=0.477}$
$\mathrm{T=298~K]}$
Explanation:
$ \begin{aligned} & \mathrm{n}_{\mathrm{NH}_3}= 0.1-0.02=0.08 \\\\ & \mathrm{n}_{\mathrm{NH}_4 \mathrm{Cl}}= \mathrm{n}_{\mathrm{NH}_4^{+}}=0.1+0.02=0.12 \\\\ & \mathrm{pOH}= \mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \\\\ &=4.745+\log \frac{0.12}{0.08} \\\\ &=4.745+\log \frac{3}{2} \\\\ &=4.745+0.477-0.301 \\\\ & \mathrm{pOH}=4.921 \\\\ & \mathrm{pH}=14-\mathrm{pH} \\\\ &=9.079 = 9079\times 10^{-3} \end{aligned} $
If the pKa of lactic acid is 5, then the pH of 0.005 M calcium lactate solution at 25$^\circ$C is ___________ $\times$ 10$^{-1}$ (Nearest integer)
Explanation:
Concentration of calcium lactate $=0.005 \mathrm{M}$, concentration of lactate ion $=(2 \times 0.005) \mathrm{M}$.
Calcium lactate is a salt of weak acid $+$ strong base
$\therefore$ Salt hydrolysis will take place.
The dissociation constant of acetic acid is $x\times10^{-5}$. When 25 mL of 0.2 $\mathrm{M~CH_3COONa}$ solution is mixed with 25 mL of 0.02 $\mathrm{M~CH_3COOH}$ solution, the pH of the resultant solution is found to be equal to 5. The value of $x$ is ____________
Explanation:
To find the dissociation constant of acetic acid, we use the Henderson-Hasselbalch equation for the given system and conditions. The equation is as follows:
$ \text{pH} = \text{pK}_\text{a} + \log \left(\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}\right) $
In the solution mixture:
We have 25 mL of 0.2 M $\text{CH}_3\text{COONa}$ and 25 mL of 0.02 M $\text{CH}_3\text{COOH}$.
The concentration ratio $\left(\frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]}\right)$ becomes $\frac{25 \times 0.2}{25 \times 0.02} = 10$.
Given that the pH of the solution is 5, we substitute into the equation:
$ 5 = \text{pK}_\text{a} + \log 10 $
Since $\log 10 = 1$, we solve for $\text{pK}_\text{a}$:
$ 5 = \text{pK}_\text{a} + 1 \quad \Rightarrow \quad \text{pK}_\text{a} = 4 $
Converting from $\text{pK}_\text{a}$ to $K_\text{a}$, we use:
$ K_\text{a} = 10^{-\text{pK}_\text{a}} = 10^{-4} $
Thus, since the dissociation constant $K_\text{a}$ is given as $x \times 10^{-5}$, compare:
$ 10^{-4} = 10 \times 10^{-5} $
Therefore, $x = 10$.
$200 \mathrm{~mL}$ of $0.01 \,\mathrm{M} \,\mathrm{HCl}$ is mixed with $400 \mathrm{~mL}$ of $0.01 \,\mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}$. The $\mathrm{pH}$ of the mixture is _________.
Given: $\log {2}=0.30, \log 3=0.48, \log 5=0.70, \log 7=0.84, \log 11=1.04$
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Permanganate titrations are not performed in presence of hydrochloric acid.
Reason R : Chlorine is formed as a consequence of oxidation of hydrochloric acid.
In the light of the above statements, choose the correct answer from the options given below
The plot of $\mathrm{pH}$-metric titration of weak base $\mathrm{NH}_{4} \mathrm{OH}$ vs strong acid HCl looks like :
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 $\mathrm{NH}_{4} \mathrm{Cl}=53.5 \mathrm{~g} \mathrm{~mol}^{-1}$ )
${K_{{a_1}}}$, ${K_{{a_2}}}$ 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 :
$20 \mathrm{~mL}$ 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 : $\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=1 \times 10^{-5}, \log 2=0.30, \log 3=0.48, \log 5=0.69, \log 7=0.84, \log 11= 1.04)$
The solubility of AgCl will be maximum in which of the following?
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 : ${K_a}(C{H_3}C{H_2}COOH) = 1.3 \times {10^{ - 5}}$
The Ksp for bismuth sulphide (Bi2S3) is 1.08 $\times$ 10$-$73. The solubility of Bi2S3 in mol L$-$1 at 298 K is :
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 :
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]
Explanation:
$\mathrm{S}=\sqrt{K_{s p}}=\sqrt{8 \times 10^{-28}}=2 \sqrt{2} \times 10^{-14}$
$=2.82 \times 10^{-14}$
$=282 \times 10^{-16}$
$ \therefore $ Ans. 282
$\mathrm{K}_{\mathrm{a}}$ 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 $\log 2=0.30$]
Explanation:
$\mathrm{pH}$ 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} $
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 : $\mathrm{CaF}_{2}=78 \mathrm{~g} \mathrm{~mol}^{-1}$)
Explanation:
$ \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} $
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 ___________.
Explanation:
During titration of oxalic acid by $\mathrm{KMnO}_{4}$, oxalic acid converts into $\mathrm{CO}_{2}$.
$\therefore$ Change in oxidation state of carbon $=1$
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 ___________.
Explanation:
$ \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} $
pH value of 0.001 M NaOH solution is ____________.
Explanation:
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
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}}$
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 :
Select correct statement from the following :
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 :
[Given : The solubility product of Ca(OH)2 in water = 5.5 $\times$ 10$-$6]
(Given : The solubility product of Zn(OH)2 is 2 $\times$ 10$-$20)
Explanation:
Due to common-ion effect (presence of NaOH) the concentration of OH$-$ 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
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
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
Explanation:
$ \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
Explanation:
${pH} = {p{ka}} + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$
$ \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$
Explanation:
$ \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$
Explanation:
$\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, ${{{S_{{A_2}X}}} \over {{S_{MX}}}} = {{{{10}^{ - 4}}} \over {2 \times {{10}^{ - 6}}}} = 50$
Explanation:
pH = 7 + ${1 \over 2}$(pka $-$ pkb)
= 7 + ${1 \over 2}$ (5.23 $-$ 4.75)
= 7.24 $ \approx $ 7.
Explanation:
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
(A) 0.01 M HCl
(B) 0.01 M NaOH
(C) 0.01 M CH3COONa
(D) 0.01 M NaCl
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.
$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 ?
Assertion : The pH of water increases with increase in temperature.
Reason : The dissociation of water into H+ and OH– is an exothermic reaction.
(Note : consider that an appropriate indicator is used)
[Assuming that neither kind of ion reacts with water]
Explanation:
| AB2 | ⇌ | A2+(aq) | + | 2B-(aq) |
|---|---|---|---|---|
| s | 2s |
Ksp = 4s3 = 3.2 × 10–11
$ \Rightarrow $ s3 = 8 × 10–12
$ \Rightarrow $ s = 2 × 10–4
(First dissociation constant of
H2CO3 = 4.0 $ \times $ 10–7; log 2 = 0.3; density
of the soft drink = 1 g mL–1) .
Explanation:
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
[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.
Explanation:
= ${3 \over {60}} \times 1000 \times {{20} \over {500}}$ = 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
Explanation:
$ \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
0.1 M Formic acid (A),
0.1 M Acetic acid (B),
0.1 M Benzoic acid (C)