Ionic Equilibrium
Consider a weak base ' B ' of $\mathrm{pK}_{\mathrm{b}}=5.699$. ' $x$ ' mL of 0.02 M HCl and ' y ' mL of 0.02 M weak base ' B ' are mixed to make 100 mL of a buffer of pH 9 at $25^{\circ} \mathrm{C}$. The values of ' $x$ ' and ' $y$ ' respectively are :
[Given : $\log 2=0.3010, \log 3=0.4771, \log 5=0.699$ ]
$ \begin{array}{|c|c|} \hline x & y \\ \hline \hline 42.7 & 57.3 \\ \hline \end{array} $
$ \begin{array}{|c|c|} \hline x & y \\ \hline \hline 14.3 & 85.7 \\ \hline \end{array} $
$ \begin{array}{|c|c|} \hline x & y \\ \hline \hline 85.7 & 14.3 \\ \hline \end{array} $
$ \begin{array}{|c|c|} \hline x & y \\ \hline \hline 11.1 & 88.9 \\ \hline \end{array} $
Which of the following mixture gives a buffer solution with $\mathrm{pH}=9.25$ ?
Given : $\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=4.75$
$0.5 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}(0.2 \mathrm{~L})+0.2 \mathrm{M} \mathrm{HCl}(0.5 \mathrm{~L})$
$0.2 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}(0.5 \mathrm{~L})+0.1 \mathrm{M} \mathrm{HCl}(0.5 \mathrm{~L})$
$0.2 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}(0.4 \mathrm{~L})+0.1 \mathrm{M} \mathrm{HCl}(1 \mathrm{~L})$
$0.4 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}(1 \mathrm{~L})+0.1 \mathrm{M} \mathrm{HCl}(1 \mathrm{~L})$
Consider the dissociation equilibrium of the following weak acid
$ \mathrm{HA} \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq}) $
If the pKa of the acid is 4 , then the pH of 10 mM HA solution is $\_\_\_\_$ .(Nearest integer)
[Given: The degree of dissociation can be neglected with respect to unity]
Explanation:
Given $pK_a = 4$, so
$K_a = 10^{-pK_a} = 10^{-4}$
For a weak acid solution of concentration $C = 10\,\text{mM} = 0.01\,\text{M}$:
Let the dissociation be $x$:
$\mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^-$
At equilibrium:
$[\mathrm{H}^+] = x$
$[\mathrm{A}^-] = x$
$[\mathrm{HA}] = C - x \approx C$ (given $x$ is negligible compared to $C$)
So,
$K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]} \approx \frac{x^2}{C}$
Therefore,
$x = \sqrt{K_a C} = \sqrt{(10^{-4})(10^{-2})} = \sqrt{10^{-6}} = 10^{-3}$
So,
$[\mathrm{H}^+] = 10^{-3}\,\text{M} \Rightarrow pH = -\log(10^{-3}) = 3$
Nearest integer = $\boxed{3}$
Consider two Group IV metal ions $\mathrm{X}^{2+}$ and $\mathrm{Y}^{2+}$.
A solution containing $0.01 \mathrm{M} \mathrm{X}^{2+}$ and $0.01 \mathrm{M} \mathrm{Y}^{2+}$ is saturated with $\mathrm{H}_2 \mathrm{~S}$. The pH at which the metal sulphide YS will form as a precipitate is $\_\_\_\_$ . (Nearest integer)
(Given: $\mathrm{K}_{\mathrm{sp}}(\mathrm{XS})=1 \times 10^{-22}$ at $25^{\circ} \mathrm{C}, \mathrm{K}_{\mathrm{sp}}(\mathrm{YS})=4 \times 10^{-16}$ at $25^{\circ} \mathrm{C}$, $\left[\mathrm{H}_2 \mathrm{~S}\right]=0.1 \mathrm{M}$ in solution, $\mathrm{K}_{a 1} \times \mathrm{K}_{a 2}\left(\mathrm{H}_2 \mathrm{~S}\right)=1.0 \times 10^{-21}, \log 2=0.30$, $\log 3=0.48, \log 5=0.70)$
Explanation:
$ \mathrm{XS}(\mathrm{~s}) \rightleftharpoons \mathrm{X}^{+2} \text { (aq.) }+\mathrm{S}^{2-} \text { (aq.) } $
For precipitation of $\mathrm{XS}(\mathrm{s})$, we use the solubility product condition:
$ \begin{aligned} & {\left[\mathrm{X}^{+2}\right]\left[\mathrm{S}^{2-}\right] \geq \mathrm{K}_{\mathrm{sp}}(\mathrm{XS})} \\ & \text{Given } [\mathrm{X}^{2+}] = 0.01 \text{ M, so } \left[\mathrm{S}^{2-}\right] \geq \frac{1 \times 10^{-22}}{0.01}=10^{-20} \\ & \mathrm{YS}(\mathrm{~s}) \rightleftharpoons \mathrm{Y}^{+2}(\mathrm{aq})+\mathrm{S}^{2-}(\mathrm{aq}) \end{aligned} $
For precipitation of $\mathrm{YS}(\mathrm{s})$, again apply $K_{sp}$ condition:
$ \begin{aligned} & {\left[\mathrm{Y}^{+2}\right]\left[\mathrm{S}^{2-}\right] \geq \mathrm{K}_{\mathrm{sp}}(\mathrm{YS})} \\ & \text{Given } [\mathrm{Y}^{2+}] = 0.01 = 10^{-2} \text{ M, so } \left[\mathrm{S}^{2-}\right] \geq \frac{4 \times 10^{-16}}{10^{-2}}=4 \times 10^{-14} \end{aligned} $
Now we find how much $\mathrm{S}^{2-}$ comes from $\mathrm{H}_2\mathrm{S}$. In water:
Now, $\mathrm{H}_2 \mathrm{~S}(\mathrm{aq}) \rightleftharpoons 2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{S}^{2-}(\mathrm{aq})$
Using the given relation $\mathrm{K}_{a1}\times \mathrm{K}_{a2}=1 \times 10^{-21}$:
$\begin{aligned} & \frac{\left[\mathrm{S}^{2-}\right]\left[\mathrm{H}^{+}\right]^2}{\mathrm{H}_2 \mathrm{~S}}=\mathrm{K}_{\mathrm{a}_1} \times \mathrm{K}_{\mathrm{a}_2}=1 \times 10^{-21} \\ & \Rightarrow {\left[\mathrm{~S}^{2-}\right]=\frac{1 \times 10^{-21} \times\left[\mathrm{H}_2 \mathrm{~S}\right]}{\left[\mathrm{H}^{+}\right]^2}} \end{aligned}$
For $\mathrm{YS}$ to start precipitating, we need $\left[\mathrm{S}^{2-}\right] \geq 4 \times 10^{-14}$. So:
$\begin{aligned} & \frac{1 \times 10^{-21} \times\left[\mathrm{H}_2 \mathrm{~S}\right]}{\left[\mathrm{H}^{+}\right]^2} \geq 4 \times 10^{-14} \\ & \text{Given } [\mathrm{H}_2\mathrm{S}] = 0.1 = 10^{-1} \text{ M} \\ & \Rightarrow \frac{1 \times 10^{-21} \times 10^{-1}}{\left[\mathrm{H}^{+}\right]^2} \geq 4 \times 10^{-14} \\ & \Rightarrow \left[\mathrm{H}^{+}\right]^2 \leq \frac{10^{-22}}{4 \times 10^{-14}}=\frac{1}{4}\times 10^{-8} \\ & { \left[\mathrm{H}^{+}\right]^2 \leq \frac{1}{4} \times 10^{-7} \times 10^{-1}} \\ & \Rightarrow { \left[\mathrm{H}^{+}\right] \leq \frac{1}{2} \times 10^{-4}} \\ & \Rightarrow \mathrm{pH} \geq 4.3 \end{aligned}$
The first and second ionization constants of H2X are $2.5 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively.
The concentration of ${X^{2-}}$ in $0.1\ \mathrm{M}$ H2X solution is ______ $\times 10^{-15}\ \mathrm{M}$. (Nearest Integer)
Explanation:
For the diprotic acid $ \mathrm{H_2X} $ (total concentration $C=0.1\,\mathrm{M}$):
$ K_{a1}=2.5\times 10^{-8},\qquad K_{a2}=1.0\times 10^{-13} $
1) Find $[\mathrm{H^+}]$ (second dissociation is negligible for $[\mathrm{H^+}]$):
$ [\mathrm{H^+}] \approx \sqrt{K_{a1}C} = \sqrt{(2.5\times 10^{-8})(0.1)} = \sqrt{2.5\times 10^{-9}} = 5.0\times 10^{-5}\,\mathrm{M} $
2) Use diprotic distribution for $[\mathrm{X^{2-}}]$:
$ [\mathrm{X^{2-}}]= C\cdot \frac{K_{a1}K_{a2}/[\mathrm{H^+}]^2}{1+K_{a1}/[\mathrm{H^+}]+K_{a1}K_{a2}/[\mathrm{H^+}]^2} $
Compute key terms:
$ \frac{K_{a1}K_{a2}}{[\mathrm{H^+}]^2} =\frac{(2.5\times 10^{-8})(10^{-13})}{(5\times 10^{-5})^2} =\frac{2.5\times 10^{-21}}{2.5\times 10^{-9}} =10^{-12} $
$ \frac{K_{a1}}{[\mathrm{H^+}]}=\frac{2.5\times 10^{-8}}{5\times 10^{-5}}=5\times 10^{-4} $
So denominator $\approx 1+5\times 10^{-4}\approx 1.0005$, hence
$ [\mathrm{X^{2-}}] \approx 0.1 \times \frac{10^{-12}}{1.0005} \approx 1.0\times 10^{-13}\,\mathrm{M} $
Expressed as $\_\_\_\_ \times 10^{-15}\,\mathrm{M}$:
$ 1.0\times 10^{-13}\,\mathrm{M} = 100\times 10^{-15}\,\mathrm{M} $
Nearest integer = $\boxed{100}$.
Given is a concentrated solution of a weak electrolyte $A_x B_y$ of concentration ' $c$ ' and dissociation constant ' K '. The degree of dissociation is given by :
$ \left[\mathrm{K} \times \mathrm{c}^{x+y-1} x^x y^y\right]^{x+y} $
$ \left(\frac{\mathrm{K}}{\mathrm{c}^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}} $
$ \left(\frac{\mathrm{c}^{x+y-1} x^x y^y}{\mathrm{~K}}\right)^{x+y} $
$ \left(\frac{c^{x+y-1} x^x y^y}{\mathrm{~K}}\right)^{\frac{1}{x+y}} $
$\mathrm{M}_3 \mathrm{~A}_2$ is a sparingly soluble salt of molar mass $y \mathrm{~g} \mathrm{~mol}^{-1}$ and solubility $x \mathrm{~g} \mathrm{~L}^{-1}$. The ratio of the molar concentration of the anion $\left(\mathrm{A}^{3-}\right)$ to the solubility product of the salt is
$\frac{1}{54} \cdot \frac{y^4}{x^4}$
$\frac{y^5}{108 x^4}$
$108 \cdot \frac{x^5}{y^5}$
$ \frac{1}{108} \frac{y^4}{x^4} $
Arrange the following resultant mixtures in increasing order of their pH values
A. $10 \mathrm{~mL} 0.2 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2+25 \mathrm{~mL} 0.1 \mathrm{M} \mathrm{HCl}$
B. $10 \mathrm{~mL} 0.01 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4+10 \mathrm{~mL} 0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2$
C. $10 \mathrm{~mL} 0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4+10 \mathrm{~mL} 0.1 \mathrm{M} \mathrm{KOH}$
Choose the correct answer from the options given below :
$\mathrm{B}<\mathrm{C}<\mathrm{A}$
$\mathrm{C}<\mathrm{A}<\mathrm{B}$
$\mathrm{C}<\mathrm{B}<\mathrm{A}$
$\mathrm{A}<\mathrm{C}<\mathrm{B}$
Given below are two statements :
Statement I : Sodium dichromate and potassium dichromate are classified as primary standards in titrimetric analysis.
Statement II : Phenolphthalein is a weak base, therefore it dissociates in acidic medium.
In the light of the above statements, choose the correct answer from the options given below
Both Statement I and Statement II are ture
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true
20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution $(\mathrm{X})$ ? $\left(\mathrm{pK}_{\mathrm{a}}\right.$ value of acetic acid is 4.75).
7.0
4.75
3.5
4.82
The first and second ionization constants of a weak dibasic acid H2A are $8.1 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. 0.1 mol of H2A was dissolved in 1 L of 0.1 M HCl solution. The concentration of HA- in the resultant solution is:
$0.1\ \mathrm{M}$
$9.53 \times 10^{-6}\ \mathrm{M}$
$8.1 \times 10^{-8}\ \mathrm{M}$
$1.0 \times 10^{-13}\ \mathrm{M}$
At 25°C, 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when 10 mL of NaOH is added respectively, are :
Given :
$K_a = 5 \times 10^{-4}$
$\mathrm{p}K_a = 3.3$
$\alpha \ll 1$
| (a) | (b) |
|---|---|
| 0.7 | 2.0 |
| (a) | (b) |
|---|---|
| 2.0 | 3.3 |
| (a) | (b) |
|---|---|
| 1.1 | 2.2 |
| (a) | (b) |
|---|---|
| 3.0 | 2.2 |
The solubility product constants of $\mathrm{Ag_2CrO_4}$ and $\mathrm{AgBr}$ are $32x$ and $4y$ respectively at 298 K.
The value of $\left( \frac{\text{molarity of } \mathrm{Ag_2CrO_4}}{\text{molarity of } \mathrm{AgBr}} \right)$ can be expressed as :
$\frac{2\sqrt[3]{x}}{y}$
$2 \sqrt{\frac{x}{y}}$
$\sqrt{\frac{x}{y}}$
$\frac{\sqrt[3]{x}}{\sqrt{y}}$
The pH of a solution obtained by mixing 5 mL of $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$ solution with 250 mL of $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{Cl}$ solution is $\_\_\_\_$ $\times 10^{-2}$.(Nearest integer) Given: $\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=4.74$
$ \begin{aligned} & \log 2=0.30 \\ & \log 3=0.48 \\ & \log 5=0.70 \end{aligned} $
Explanation:
For the mixture of a weak base and its salt, we use the buffer formula for a basic buffer:
$ \mathrm{pOH}=\mathrm{p}K_b+\log \frac{[\text{salt}]}{[\text{base}]} $
Here,
Weak base $= \mathrm{NH_4OH}$
Salt $= \mathrm{NH_4Cl}$
Since both are mixed in the same final solution, we can use moles directly.
Step 1: Calculate moles of base
Volume of $\mathrm{NH_4OH}=5\,\text{mL}=0.005\,\text{L}$
Molarity $=0.1\,\text{M}$
$ n(\mathrm{NH_4OH})=0.1 \times 0.005=5\times 10^{-4}\,\text{mol} $
Step 2: Calculate moles of salt
Volume of $\mathrm{NH_4Cl}=250\,\text{mL}=0.250\,\text{L}$
Molarity $=0.1\,\text{M}$
$ n(\mathrm{NH_4Cl})=0.1 \times 0.250=2.5\times 10^{-2}\,\text{mol} $
Step 3: Apply buffer formula
$ \mathrm{pOH}=4.74+\log \frac{2.5\times 10^{-2}}{5\times 10^{-4}} $
$ \frac{2.5\times 10^{-2}}{5\times 10^{-4}}=50 $
So,
$ \mathrm{pOH}=4.74+\log 50 $
Now,
$ \log 50=\log(5\times 10)=\log 5 +1 =0.70+1=1.70 $
Hence,
$ \mathrm{pOH}=4.74+1.70=6.44 $
Step 4: Find pH
$ \mathrm{pH}=14-6.44=7.56 $
Now write it in the form asked:
$ 7.56 = 756 \times 10^{-2} $
Nearest integer $=756$
Final Answer:
$ \boxed{756} $
An aqueous solution of HCl with pH 1.0 is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would
$($ Given $\log 2=0.30)$
40 mL of a mixture of $\mathrm{CH}_3 \mathrm{COOH}$ and HCl (aqueous solution) is titrated against 0.1 M NaOH solution conductometrically. Which of the following statement is correct?
If equal volumes of $A B_2$ and $X Y$ (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of $\mathrm{AY}_2$ at 300 K ? (Given $\mathrm{K}_{\mathrm{sp}}\left(\right.$ at 300 K ) for $\mathrm{AY}_2=5.2 \times 10^{-7}$ )
Arrange the following in increasing order of solubility product :
$\mathrm{Ca}(\mathrm{OH})_2, \mathrm{AgBr}, \mathrm{PbS}, \mathrm{HgS}$
A weak acid HA has degree of dissociation x . Which option gives the correct expression of ( pH - $\mathrm{pK}_{\mathrm{a}}$)?
$\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{Cr}(\mathrm{OH})_3$ is $1.6 \times 10^{-30}$. What is the molar solubility of this salt in water?
pH of water is 7 at $25^{\circ} \mathrm{C}$. If water is heated to $80^{\circ} \mathrm{C}$., it's pH will :
Which of the following happens when $\mathrm{NH}_4 \mathrm{OH}$ is added gradually to the solution containing 1 M $\mathrm{A}^{2+}$ and $1 \mathrm{M} \mathrm{B}^{3+}$ ions?
Given : $\mathrm{K}_{\text {sp }}\left[\mathrm{A}(\mathrm{OH})_2\right]=9 \times 10^{-10}$ and $\mathrm{K}_{\mathrm{sp}}\left[\mathrm{B}(\mathrm{OH})_3\right]=27 \times 10^{-18}$ at 298 K.
The molar solubility(s) of zirconium phosphate with molecular formula $\left(\mathrm{Zr}^{4+}\right)_3\left(\mathrm{PO}_4^{3-}\right)_4$ is given by relation :
$\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=(3 \mathrm{~s})^3 \cdot(4 \mathrm{~s})^4 \\\\ & \mathrm{~K}_{\mathrm{sp}}=6912 \mathrm{~s}^7 \\\\ & \mathrm{~s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{7}}\end{aligned}$
One litre buffer solution was prepared by adding 0.10 mol each of $\mathrm{NH}_3$ and $\mathrm{NH}_4 \mathrm{Cl}$ in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is ______________ $\times 10^{-2}$.
(Nearest integer)
Given : $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{NH}_3=4.745$ and $\log _{10} 3=0.477$
Explanation:
$\begin{aligned} &\begin{aligned} & \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \\ & \mathbf{p O H}=4.745 \end{aligned}\\ &\text { on adding } 0.05 \text { mole } \mathrm{HCl} \end{aligned}$
$\begin{array}{llll} \mathrm{NH}_3+ & \mathrm{H}^{\oplus} & \rightarrow & \mathrm{NH}_4^{\oplus} \\ 0.1 & 0.05 & & 0.1 \\ 0.05 & 0 & & 0.15 \end{array}$
$\begin{aligned} & \mathrm{pOH}^{\prime}=4.745+\log 3 \\ & \mathrm{pOH}^{\prime}-\mathrm{pOH}=0.477 \\ & 14-\mathrm{pH}^{\prime}-14+\mathrm{pH}=0.477 \\ & \Delta \mathrm{pH}=0.477 \\ & =47.7 \times 10^{-2} \approx 48 \times 10^{-2} \end{aligned}$
The percentage dissociation of a salt $\left(\mathrm{MX}_3\right)$ solution at given temperature (van't Hoff factor $\mathrm{i}=2$ ) is ___________ %(Nearest integer)
Explanation:
To determine the percentage dissociation of the salt $\text{MX}_3$, consider the following dissociation reaction:
$ \text{MX}_3 \rightarrow \text{M}^{+3} + 3\text{X}^{-} $
The van't Hoff factor, $\text{i}$, is given as 2. The formula relating $\text{i}$ to the degree of dissociation, $\alpha$, is:
$ \text{i} = 1 + (\text{n} - 1) \alpha $
Here, $\text{n}$ represents the total number of ions produced per formula unit of the salt. For $\text{MX}_3$, dissociation yields:
1 $\text{M}^{+3}$ ion
3 $\text{X}^{-}$ ions
Thus, $\text{n} = 4$. Substituting into the formula, we have:
$ 2 = 1 + (4 - 1) \alpha $
Simplifying, we solve for $\alpha$:
$ 2 = 1 + 3\alpha $
$ 3\alpha = 1 $
$ \alpha = \frac{1}{3} \approx 0.3333 $
As a percentage, this degree of dissociation is:
$ \alpha \times 100\% \approx 33.33\% $
Rounding to the nearest integer gives:
$ 33\% $
Therefore, the percentage dissociation of the salt $\text{MX}_3$ is approximately 33%.
The molar conductance of an infinitely dilute solution of ammonium chloride was found to be $185 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and $70 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is $85.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, its degree of dissociation is given by $x \times 10^{-1}$. The value of $x$ is __________ . (Nearest integer)
Explanation:
$\begin{aligned} & \lambda_{\mathrm{m}}^{\circ} \text { of } \mathrm{NH}_4 \mathrm{Cl}=185 \\ & \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4^{+}}+\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Cl}^{-}}=185 \\ & \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4^{+}}=185-70=115 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\ & \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}=\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4^{+}}+\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{OH}^{-}} \\ & =115+170 \\ & \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}=285 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \end{aligned}$
$\begin{aligned} \text { degree of dissociation } & =\frac{\left(\lambda_{\mathrm{m}}\right)_{\mathrm{NH}_4 \mathrm{OH}}}{\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}} \\ & =\frac{85.5}{285} \\ & =0.3 \\ & =3 \times 10^{-1} \end{aligned}$
$x \mathrm{mg}$ of $\mathrm{Mg}(\mathrm{OH})_2($ molar mass $=58)$ is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K . The value of $x$ is ________ mg. (Nearest integer)
(Given : $\mathrm{Mg}(\mathrm{OH})_2$ is assumed to dissociate completely in $\mathrm{H}_2 \mathrm{O}$ ]
Explanation:
To determine the mass of Mg(OH)₂ that needs to be dissolved to achieve a pH of 10.0, follow these steps:
Given:
pH = 10
Therefore, pOH = 14 - pH = 4
[OH⁻] = 10⁻⁴ M
First, calculate the number of moles of OH⁻ ions:
Number of moles of OH⁻ = 10⁻⁴
Since Mg(OH)₂ dissociates completely in water, from one mole of Mg(OH)₂, you get two moles of OH⁻:
Number of moles of Mg(OH)₂ = (10⁻⁴) / 2 = 5 × 10⁻⁵ moles
Next, calculate the mass of Mg(OH)₂:
Molar mass of Mg(OH)₂ = 58 g/mol
Convert the number of moles to mass:
Mass of Mg(OH)₂ = (5 × 10⁻⁵ moles) × (58 g/mol) = 2.9 × 10⁻³ g
Convert to milligrams: 2.9 × 10⁻³ g = 2.9 mg
Therefore, the calculated mass of Mg(OH)₂ required is approximately 2.9 mg.
The pH of a 0.01 M weak acid $\mathrm{HX}\left(\mathrm{K}_a=4 \times 10^{-10}\right)$ is found to be 5 . Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6 . The new concentration of the diluted weak acid is given as $x \times 10^{-4} \mathrm{M}$. The value of $x$ is _________ (nearest integer)
Explanation:
$\mathrm{HX}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-} \mathrm{K}_{\mathrm{a}}=4 \times 10^{-10}$
$\begin{aligned} &0.01(1-\alpha) \quad 0.01 \alpha \quad 0.01 \alpha \quad \text { Not justified }\\ &\begin{aligned} & \Rightarrow 0.01 \alpha=10^{-5} \Rightarrow \alpha=10^{-3} \\ & \mathrm{~K}_{\mathrm{a}}=0.01 \alpha^2=10^{-8} \end{aligned} \end{aligned}$
$\begin{aligned} &\text { On dilution let final concentration of } \mathrm{HX}=\mathrm{c} \mathrm{M}\\ &\mathrm{Hx}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-} \end{aligned}$
$\mathrm{C}(1-\alpha) \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha$
$\begin{aligned} \Rightarrow & C \alpha=10^{-6} \quad\text{..... (1)}\\ & \frac{C \alpha^2}{1-\alpha}=\mathrm{K}_{\mathrm{a}}=10^{-8} \quad\text{..... (2)}\\ \Rightarrow & \frac{10^{-6} \alpha}{1-\alpha}=10^{-8} \end{aligned}$
Data given is inconsistent & contradictory. This should be bonus.
The observed and normal molar masses of compound $\mathrm{MX}_2$ are 65.6 and 164 respectively. The percent degree of ionisation of $\mathrm{MX}_2$ is __________%. (Nearest integer)
Explanation:
To calculate the percent degree of ionization of the compound $\mathrm{MX}_2$, we begin by considering its dissociation process:
$ \mathrm{MX}_2 \rightarrow \mathrm{M}^{+2} + 2\mathrm{X}^{-} $
Next, we use the van't Hoff factor (i), calculated as the ratio of the normal molar mass to the observed molar mass:
$ i = \frac{\text{normal molar mass}}{\text{observed molar mass}} = \frac{164}{65.6} $
Now, expressing the relationship of ionization in terms of the van't Hoff factor, we have:
$ 1 + (3 - 1)\alpha = \frac{164}{65.6} $
Simplifying this equation gives us:
$ 2\alpha = \frac{164}{65.6} - 1 = \frac{98.4}{65.6} $
Solving for $\alpha$:
$ \alpha = 0.75 $
Therefore, the percent dissociation, or the percent degree of ionization, is:
$ \text{Percent dissociation} = 75\% $
If 1 mM solution of ethylamine produces $\mathrm{pH}=9$, then the ionization constant $\left(\mathrm{K}_{\mathrm{b}}\right)$ of ethylamine is $10^{-x}$. The value of $x$ is _________ (nearest integer).
[The degree of ionization of ethylamine can be neglected with respect to unity.]
Explanation:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{C}_2 \mathrm{H}_2 \mathrm{NH}_3^{+}+\stackrel{\ominus}{\mathrm{O}} \mathrm{H}$
$\begin{array}{ccc} \mathrm{C}=10^{-3} \mathrm{M} & - & - \\ \mathrm{C}(1-\alpha) & \mathrm{C} \alpha & \mathrm{C} \alpha \\ \Rightarrow \mathrm{C}=10^{-3} & =10^{-5} & =10^{-5} \end{array}$
$1-\alpha \simeq 1$
Given, $\mathrm{P}^{\mathrm{H}}=9 \Rightarrow \mathrm{P}^{\mathrm{OH}}=5 \Rightarrow[\stackrel{\ominus}{\mathrm{O}} \mathrm{H}]=10^{-5} \mathrm{M}$
Now, $K_b=\frac{\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_3^{+}\right][\stackrel{\ominus}{\mathrm{O}} \mathrm{H}]}{\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\right]}$
$\Rightarrow \mathrm{K}_{\mathrm{b}}=\frac{10^{-5} \times 10^{-5}}{10^{-3}}=10^{-7}$
For a sparingly soluble salt $\mathrm{AB}_2$, the equilibrium concentrations of $\mathrm{A}^{2+}$ ions and $B^{-}$ ions are $1.2 \times 10^{-4} \mathrm{M}$ and $0.24 \times 10^{-3} \mathrm{M}$, respectively. The solubility product of $\mathrm{AB}_2$ is :
Given below are two statements :
Statement (I) : A Buffer solution is the mixture of a salt and an acid or a base mixed in any particular quantities
Statement (II) : Blood is naturally occurring buffer solution whose $\mathrm{pH}$ is maintained by $\mathrm{H}_2 \mathrm{CO}_3 / \mathrm{HCO}_3{ }^{\ominus}$ concentrations.
In the light of the above statements, choose the correct answer from the options given below :
The equilibrium $\mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightleftharpoons 2 \mathrm{CrO}_4^{2-}$ is shifted to the right in :
Given below are two statements :
Statement (I) : Aqueous solution of ammonium carbonate is basic.
Statement (II) : Acidic/basic nature of salt solution of a salt of weak acid and weak base depends on $K_a$ and $K_b$ value of acid and the base forming it.
In the light of the above statements, choose the most appropriate answer from the options given below :
Consider the dissociation of the weak acid HX as given below
$\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}$
[$\mathrm{K}_{\mathrm{a}}$ : dissociation constant]
The osmotic pressure of $0.03 \mathrm{M}$ aqueous solution of $\mathrm{HX}$ at $300 \mathrm{~K}$ is _________ $\times 10^{-2}$ bar (nearest integer).
[Given : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$]
Explanation:
$\begin{aligned} & \mathrm{Ka}=\frac{\mathrm{C} \alpha^2}{1-\alpha} \\ & 1.2 \times 10^{-5}=(0.03)\left(\alpha^2\right) \\ & \alpha=0.02 \\ & \pi=\mathrm{iCRT} \\ &=(1.02)(0.03)(0.083)(300) \\ &=0.76194 \\ &=76.194 \times 10^{-2} \mathrm{bar} \\ & \text { Nearest integer }=76 \end{aligned}$
Explanation:
To find the pH of an ammonium acetate solution, we use the fact that ammonium acetate is a salt resulting from the neutralization of a weak acid (acetic acid, $CH_3COOH$) by a weak base (ammonium hydroxide, $NH_4OH$). The respective ionization constant values for the weak acid $K_a$ and the weak base $K_b$ are given as $1.8 \times 10^{-5}$.
Firstly, calculate the $pK_a$ and $pK_b$ values.
- $pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5})$
- $pK_b = -\log(K_b) = -\log(1.8 \times 10^{-5})$
Given that the values of $K_a$ and $K_b$ are the same, their $pK_a$ and $pK_b$ values will also be the same, establishing a neutral condition where the effects of the acid and base neutralize each other.
The formula used to determine the pH of a solution of such a salt is:
$pH = \frac{1}{2}(pK_w + pK_a - pK_b)$
Where $pK_w$ is the ionic product of water, which is 14 at 25°C. In this specific case, since $pK_a = pK_b$, the formula simplifies to:
$pH = \frac{1}{2}(14 + pK_a - pK_a)$
$pH = \frac{1}{2}(14)$
$pH = 7$
The pH of an ammonium acetate solution in this scenario is 7, indicating a neutral solution.
The $\mathrm{pH}$ of an aqueous solution containing $1 \mathrm{M}$ benzoic acid $\left(\mathrm{pK}_{\mathrm{a}}=4.20\right)$ and $1 \mathrm{M}$ sodium benzoate is 4.5. The volume of benzoic acid solution in $300 \mathrm{~mL}$ of this buffer solution is _________ $\mathrm{mL}$. (given : $\log 2=0.3$)
Explanation:
$\begin{array}{ccc} & \text { 1M Benzoic acid } & +1 \mathrm{M} \text { Sodium Benzoate } \\ & \left(\mathrm{V}_{\mathrm{a}} \mathrm{ml}\right) & \left(\mathrm{V}_{\mathrm{s}} \mathrm{ml}\right) \\ \text { Millimole } & \mathrm{V}_{\mathrm{a}} \times 1 & \mathrm{~V}_{\mathrm{s}} \times 1 \end{array}$
$\begin{aligned} & \mathrm{pH}=4.5 \\ & \mathrm{pH}=\mathrm{pka}+\log \frac{[\text { salt }]}{[\text { acid }]} \\ & 4.5=4.2+\log \left(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{a}}}\right) \\ & \frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{a}}}=2 \quad \text{..... (1)}\\ & \mathrm{~V}_{\mathrm{s}}+\mathrm{V}_{\mathrm{a}}=300 \quad \text{..... (2)}\\ & \mathrm{~V}_{\mathrm{a}}=100 \mathrm{~ml} \end{aligned}$
The $\mathrm{pH}$ at which $\mathrm{Mg}(\mathrm{OH})_2\left[\mathrm{~K}_{\mathrm{sp}}=1 \times 10^{-11}\right]$ begins to precipitate from a solution containing $0.10 \mathrm{~M} \mathrm{~Mg}^{2+}$ ions is __________.
Explanation:
$\begin{aligned} & \text { Precipitation when } Q_{s p}=K_{s p} \\ & {\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2=10^{-11}} \\ & 0.1 \times\left[\mathrm{OH}^{-}\right]^2=10^{-11} \Rightarrow\left[\mathrm{OH}^{-}\right]=10^{-5} \\ & \Rightarrow \mathrm{pOH}=5 \quad \Rightarrow \mathrm{pH}=9 \end{aligned}$
(A) The $\mathrm{pH}$ of $1 \times 10^{-8}~ \mathrm{M} ~\mathrm{HCl}$ solution is 8 .
(B) The conjugate base of $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$ is $\mathrm{HPO}_{4}^{2-}$.
(C) $\mathrm{K}_{\mathrm{w}}$ increases with increase in temperature.
(D) When a solution of a weak monoprotic acid is titrated against a strong base at half neutralisation point, $\mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}$
Choose the correct answer from the options given below:
$25 \mathrm{~mL}$ of silver nitrate solution (1M) is added dropwise to $25 \mathrm{~mL}$ of potassium iodide $(1.05 \mathrm{M})$ solution. The ion(s) present in very small quantity in the solution is/are :
When the hydrogen ion concentration [H$^+$] changes by a factor of 1000, the value of pH of the solution __________
20 mL of $0.1 ~\mathrm{M} ~\mathrm{NaOH}$ is added to $50 \mathrm{~mL}$ of $0.1 ~\mathrm{M}$ acetic acid solution. The $\mathrm{pH}$ of the resulting solution is ___________ $\times 10^{-2}$ (Nearest integer)
Given : $\mathrm{pKa}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=4.76$
$\log 2=0.30$
$\log 3=0.48$
Explanation:
Moles of NaOH = Volume × Molarity = 20 mL × 0.1 M = 2 mmol
Moles of acetic acid = Volume × Molarity = 50 mL × 0.1 M = 5 mmol
Since NaOH is a strong base, it will react with acetic acid to form acetate ions (CH₃COO⁻) and water:
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
2 mmol of NaOH will react with 2 mmol of acetic acid, resulting in 2 mmol of acetate ions and leaving 3 mmol of acetic acid unreacted.
Next, we need to find the concentrations of acetic acid and acetate ions in the resulting 70 mL solution:
Concentration of acetic acid = Moles / Total volume = 3 mmol / 70 mL = 0.04286 M
Concentration of acetate ions = Moles / Total volume = 2 mmol / 70 mL = 0.02857 M
Now we can use the Henderson-Hasselbalch equation to find the pH of the resulting solution:
pH = pKa + log ([A⁻] / [HA])
Given the pKa of acetic acid is 4.76, we can substitute the values:
pH = 4.76 + log (0.02857 / 0.04286)
Using the given log values, we can approximate the log value:
log (0.02857 / 0.04286) ≈ log (2/3) ≈ log 2 - log 3 ≈ 0.30 - 0.48 = -0.18
Now substitute this value back into the Henderson-Hasselbalch equation:
pH = 4.76 - 0.18 = 4.58
To express the pH as the nearest integer multiplied by 10⁻², multiply the pH value by 100:
4.58 × 100 = 458
So, the pH of the resulting solution is approximately 458 × 10⁻².
$25.0 \mathrm{~mL}$ of $0.050 ~\mathrm{M} ~\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$ is mixed with $25.0 \mathrm{~mL}$ of $0.020 ~\mathrm{M} ~\mathrm{NaF} . \mathrm{K}_{\mathrm{Sp}}$ of $\mathrm{BaF}_{2}$ is $0.5 \times 10^{-6}$ at $298 \mathrm{~K}$. The ratio of $\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2}$ and $\mathrm{K}_{\mathrm{sp}}$ is ___________.
(Nearest integer)
Explanation:
$[\mathrm{Ba(NO_3)_2}] = 0.050\, \mathrm{M}$
$[\mathrm{NaF}] = 0.020\, \mathrm{M}$
Volumes of the solutions:
$V_{\mathrm{Ba(NO_3)_2}} = V_{\mathrm{NaF}} = 25.0\, \mathrm{mL}$
After mixing, the total volume becomes:
$V_{\mathrm{total}} = 25.0\, \mathrm{mL} + 25.0\, \mathrm{mL} = 50.0\, \mathrm{mL}$
Now, we calculate the initial concentrations after mixing:
$[\mathrm{Ba}^{2+}] = \frac{25.0\, \mathrm{mL} \times 0.050\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.025\, \mathrm{M}$
$[\mathrm{F}^{-}] = \frac{25.0\, \mathrm{mL} \times 0.020\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.010\, \mathrm{M}$
Next, we calculate the reaction quotient (Q) for the precipitation of BaF₂:
$Q = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$
$Q = (0.025\, \mathrm{M})(0.010\, \mathrm{M})^2 = 2.5 \times 10^{-6}$
Ksp of BaF₂ is given as $5 \times 10^{-7}$.
Now, we find the ratio of $[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$ to Ksp:
$\text{Ratio} = \frac{(2.5 \times 10^{-6})}{(5 \times 10^{-7})} = 5$
So, the correct ratio of $[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$ to Ksp is 5.
An analyst wants to convert $1 \mathrm{~L} \mathrm{~HCl}$ of $\mathrm{pH}=1$ to a solution of $\mathrm{HCl}$ of $\mathrm{pH} ~2$. The volume of water needed to do this dilution is __________ $\mathrm{mL}$. (Nearest integer)
Explanation:
We can use the formula for pH to calculate the concentration of hydrogen ions in each solution:
$\mathrm{pH} = -\log_{10}[\mathrm{H}^+]$
For the first solution, we have:
$1 = -\log_{10}[\mathrm{H}^+]$
Solving for $[\mathrm{H}^+]$, we get:
$[\mathrm{H}^+] = 0.1 \mathrm{~M}$
For the second solution, we want:
$2 = -\log_{10}[\mathrm{H}^+]$
Solving for $[\mathrm{H}^+]$, we get:
$[\mathrm{H}^+] = 0.01 \mathrm{~M}$
To dilute the first solution to the desired concentration, we can use the dilution equation:
$C_1V_1 = C_2V_2$
where $C_1$ and $V_1$ are the initial concentration and volume, and $C_2$ and $V_2$ are the final concentration and volume.
We know $C_1 = 0.1 \mathrm{~M}$, $C_2 = 0.01 \mathrm{~M}$, and $V_1 = 1 \mathrm{~L}$. Solving for $V_2$, we get:
$V_2 = \frac{C_1V_1}{C_2} = \frac{(0.1 \mathrm{~M})(1 \mathrm{~L})}{0.01 \mathrm{~M}} = 10 \mathrm{~L}$
Therefore, we need to add $10-1=9$ liters of water to the initial solution to obtain the desired pH. Converting liters to milliliters, we get:
$9 \mathrm{~L} \times \frac{1000 \mathrm{~mL}}{1 \mathrm{~L}} = 9000 \mathrm{~mL}$
So the volume of water needed is 9000 mL or 9,000 mL (nearest integer).
The solubility product of $\mathrm{BaSO}_{4}$ is $1 \times 10^{-10}$ at $298 \mathrm{~K}$. The solubility of $\mathrm{BaSO}_{4}$ in $0.1 ~\mathrm{M} ~\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})$ solution is ___________ $\times 10^{-9} \mathrm{~g} \mathrm{~L}^{-1}$ (nearest integer).
Given: Molar mass of $\mathrm{BaSO}_{4}$ is $233 \mathrm{~g} \mathrm{~mol}^{-1}$
Explanation:
Barium sulfate, BaSO4, is a sparingly soluble salt. Its dissolution can be represented by the following equilibrium reaction:
$\mathrm{BaSO}_4 (\mathrm{s}) \rightleftharpoons \mathrm{Ba}^{2+}(\mathrm{aq}) + \mathrm{SO}_4^{2-}(\mathrm{aq})$
The solubility product constant, Ksp, is given by:
$\mathrm{K}_{sp} = [\mathrm{Ba}^{2+}][\mathrm{SO}_4^{2-}]$
In this case, the salt is being dissolved in a solution that already contains sulfate ions, $\mathrm{SO}_4^{2-}$, from the $\mathrm{K}_2\mathrm{SO}_4$.
When $\mathrm{K}_2\mathrm{SO}_4$ dissolves completely in water, it forms $\mathrm{K}^+$ and $\mathrm{SO}_4^{2-}$ ions. Because its concentration is 0.1 M, the $\mathrm{SO}_4^{2-}$ concentration due to $\mathrm{K}_2\mathrm{SO}_4$ is 0.1 M.
Therefore, the concentration of $\mathrm{SO}_4^{2-}$ ions is now not just due to the BaSO4 dissolving, but also the added $\mathrm{K}_2\mathrm{SO}_4$. Hence, the total concentration of $\mathrm{SO}_4^{2-}$ ions is $\mathrm{S} + 0.1$ where S is the solubility of $\mathrm{BaSO}_4$.
We then substitute into the Ksp expression:
$1 \times 10^{-10} = [\mathrm{S}][\mathrm{S} + 0.1]$
However, because BaSO4 is sparingly soluble, S is very small compared to 0.1. Therefore, we can make the approximation that $\mathrm{S} + 0.1$ is approximately 0.1. This simplifies the equation to:
$1 \times 10^{-10} = 0.1[\mathrm{S}]$
Solving for S (the molar solubility of BaSO4) gives:
$S = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9} \, \mathrm{mol/L}$
To convert this molarity to a mass per volume concentration, we multiply by the molar mass of BaSO4, which is 233 g/mol:
$\mathrm{Solubility} = S \times \text{Molar mass of BaSO4} = 1 \times 10^{-9} \, \mathrm{mol/L} \times 233 \, \mathrm{g/mol} = 233 \times 10^{-9} \, \mathrm{g/L}$
So, the solubility of BaSO4 in a 0.1 M K2SO4 solution is 233 x $10^{-9}$ g/L, or 233 ng/L.
The titration curve of weak acid vs. strong base with phenolphthalein as indictor) is shown below. The $\mathrm{K}_{\text {phenolphthalein }}=4 \times 10^{-10}$.
Given: $\log 2=0.3$
The number of following statement/s which is/are correct about phenolphthalein is ___________
A. It can be used as an indicator for the titration of weak acid with weak base.
B. It begins to change colour at $\mathrm{pH}=8.4$
C. It is a weak organic base
D. It is colourless in acidic medium
Explanation:
A. This statement is incorrect. Phenolphthalein cannot be used as an indicator for the titration of a weak acid with a weak base because the pH at the equivalence point is not within the color change range of phenolphthalein (8.2-10.0).
B. This statement is correct. Phenolphthalein begins to change color at a pH of around 8.2, and the transition completes around pH 10.0. It transitions from colorless to pink as the pH increases past this point.
C. This statement is incorrect. Phenolphthalein is not a base; it is a pH indicator, which can donate protons (like an acid) but not accept them (like a base).
D. This statement is correct. Phenolphthalein is indeed colorless in acidic medium (pH less than 7), and it turns pink in a basic solution (pH greater than 7).
So, the number of correct statements is 2.
(Given mass of $\mathrm{Ag}$ is $107.9 \mathrm{~g} \mathrm{~mol}^{-1}$ and mass of $\mathrm{Cl}$ is $35.5 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Explanation:
$ \begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left(\mathrm{Ag}^{+}\right)\left(\mathrm{Cl}^{-}\right)=\mathrm{S} \times \mathrm{S}=\mathrm{S}^2 \\\\ & \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}} \\\\ & \mathrm{S}=\frac{1.434 \times 10^{-3}}{143.4}=10^{-5} \\\\ & \mathrm{~K}_{\mathrm{sp}}=\mathrm{S}^2=10^{-10} \\\\ & \Rightarrow-\log \left(\mathrm{K}_{\mathrm{sp}}\right)=-\log \left(10^{-10}\right)=10 \end{aligned} $
$600 \mathrm{~mL}$ 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 ___________ $\times 10^{-2}$. (Nearest integer)
[Given $\log 2=0.30$
$\log 3=0.48$
$\log 5=0.69$
$\log 7=0.84$
$\log 11=1.04]$
Explanation:
$\mathrm{[H^+]=\frac{6+8}{1000}=14\times10^{-3}}$
$\mathrm{pH=3-\log14}$
$=3-.3-.84$
$=1.86=186\times10^{-2}$