#

Multiple Choice Questions

One and More than One Correct

1. 2.5 mL of $\frac{2}{5}\text{ M}$ weak monoacid base ($K_b = 1 \times 10^{-12}$ at $25^\circ\text{C}$) is titrated with $\frac{2}{15}\text{ M HCl}$ in water at $25^\circ\text{C}$. The concentration of $\text{H}^+$ at equivalence point is: ($K_w = 1 \times 10^{-14}$ at $25^\circ\text{C}$) [JEE 2008 P-I]

(a) $3.7 \times 10^{-13}\text{ M}$

(b) $3.2 \times 10^{-7}\text{ M}$

(d) $2.7 \times 10^{-2}\text{ M}$

Answer: (d)

Explanation: Volume of $\text{HCl}$ used $= \frac{2.5 \times (2/5)}{2/15} = 7.5\text{ mL}$

Concentration of Salt, $C = \frac{2.5 \times (2/5)}{10} = 0.1\text{ M}$

$$\text{BOH} + \text{HCl} \longrightarrow \text{BCl} + \text{H}_2\text{O}$$

$$\text{B}^+ + \text{H}_2\text{O} \rightleftharpoons \text{BOH} + \text{H}^+$$ $$C(1-h) \quad Ch \quad Ch$$

$$\frac{C h^2}{1-h} = \frac{K_w}{K_b}$$

Solving $h = 0.27$

$$\text{∴ } [\text{H}^+] = C h = 0.1 \times 0.27 = 2.7 \times 10^{-2}\text{ M}$$

2. Solubility product constant ($K_{sp}$) of salts of types $\text{MX}$, $\text{MX}_2$ and $\text{M}_3\text{X}$ at temperature ‘T’ are $4.0 \times 10^{-8}$, $3.2 \times 10^{-14}$ and $2.7 \times 10^{-15}$, respectively. Solubilities (mole of the salts at temperature ‘T’ are in the order. [JEE 2008 P-II]

(a) $\text{MX} > \text{MX}_2 > \text{M}_3\text{X}$

(b) $\text{M}_3\text{X} > \text{MX}_2 > \text{MX}$

(d) $\text{MX} > \text{M}_3\text{X} > \text{MX}_2$

Answer: (d)

Explanation: Solubility of $(\text{MX}) = \sqrt{4 \times 10^{-8}} = 2 \times 10^{-4}\text{ mol L}^{-1}$

Solubility of $(\text{MX}_2) = \left(\frac{3.2 \times 10^{-14}}{4}\right)^{1/3} = 2 \times 10^{-5}\text{ mol L}^{-1}$

Solubility of $(\text{M}_3\text{X}) = \left(\frac{2.7 \times 10^{-15}}{27}\right)^{1/4} = 1 \times 10^{-4}\text{ mol L}^{-1}$

$$\text{∴ } \text{MX} > \text{M}_3\text{X} > \text{MX}_2$$

3. Aqueous solutions of $\text{HNO}_3$, $\text{KOH}$, $\text{CH}_3\text{COOH}$, and $\text{CH}_3\text{COONa}$ of identical concentrations are provided. The pair (s) of solutions which form a buffer upon mixing is (are): [JEE 2010 P-I]

(a) $\text{HNO}_3$ and $\text{CH}_3\text{COOH}$

(b) $\text{KOH}$ and $\text{CH}_3\text{COONa}$

(d) $\text{CH}_3\text{COOH}$ and $\text{CH}_3\text{COONa}$

Answer: (c, d)

Explanation: In option (c), if $\text{HNO}_3$ is present in limiting amount then this mixture will form a weak acid ($\text{CH}_3\text{COOH}$) and its salt ($\text{CH}_3\text{COONa}$), forming an acidic buffer. The mixture given in option (d) directly contains a weak acid and its salt with a strong base ($\text{NaOH}$), i.e., an acidic buffer.

4. The $K_{sp}$ of $\text{Ag}_2\text{CrO}_4$ is $1.1 \times 10^{-12}$ at $298\text{K}$. The solubility (in mol/L) of $\text{Ag}_2\text{CrO}_4$ in a $0.1\text{M AgNO}_3$ solution is: [JEE Adv. 2013 P-II]

(a) $1.1 \times 10^{-11}$

(b) $1.1 \times 10^{-10}$

(d) $1.1 \times 10^{-9}$

Answer: (b)

Explanation: $K_{sp} \text{ Ag}_2\text{CrO}_4 = 1.1 \times 10^{-12} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}]$

$$1.1 \times 10^{-12} = [0.1]^2 [s]$$

$$s = 1.1 \times 10^{-10}\text{ M}$$

5. $\text{p}K_a$ of a weak acid $[\text{HA}]$ and $\text{p}K_b$ of a weak base $[\text{BOH}]$ are 3.2 and 3.4, respectively. The pH of their salt $(\text{AB})$ solution is: [JEE Main 2017]

(a) 1.0

(b) 7.2

(d) 7.0

Answer: (c)

Explanation: Given $\text{p}K_a (\text{HA}) = 3.2$, $\text{p}K_b (\text{BOH}) = 3.4$

As given salt is of weak acid and weak base:

$$\text{pH} = 7 + \frac{1}{2} \text{p}K_a - \frac{1}{2} \text{p}K_b$$

$$\text{pH} = 7 + \frac{1}{2} (3.2) - \frac{1}{2} (3.4) = 6.9$$

6. An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination? [JEE Main 2018]

Base | Acid | Endpoint

(a) Strong | Strong | pink to colourless

(b) Weak | Strong | Colourless to pink

(d) Weak | Strong | Yellow to pinkish red

Answer: (d)

Explanation: The pH range of methyl orange is 3.1 to 4.4 (yellow in basic medium, pinkish red in acidic medium). Weak base is having pH greater than 7. When methyl orange is added to weak base solution, the solution becomes yellow. This solution is titrated by strong acid and at the end point pH will be less than 3.1. Therefore solution becomes pinkish red.

7. Which of the following are Lewis acids? [JEE Main 2018]

(a) $\text{BCl}_3$ and $\text{AlCl}_3$

(b) $\text{PH}_3$ and $\text{BCl}_3$

(d) $\text{PH}_3$ and $\text{SiCl}_4$

Answer: (a)

Explanation: $\text{BCl}_3$ – electron deficient, incomplete octet; $\text{AlCl}_3$ – electron deficient, incomplete octet. Both act as Lewis acids. $\text{SiCl}_4$ can also accept a lone pair of electrons into the d-orbital of silicon, hence both option (a) and (c) are considered valid contextually, with (a) being the primary choice.

8. Which of the following salts is the most basic in aqueous solution? [JEE Main 2018]

(a) $\text{Pb(CH}_3\text{COO)}_2$

(b) $\text{Al(CN)}_3$

(d) $\text{FeCl}_3$

Answer: (c)

Explanation: $\text{CH}_3\text{COOK} + \text{H}_2\text{O} \longrightarrow \text{CH}_3\text{COOH} + \text{KOH}$ (salt of weak acid and strong base $\longrightarrow$ basic solution).

$\text{FeCl}_3$ gives acidic solution. $\text{Al(CN)}_3$ is a salt of weak acid and weak base. $\text{CH}_3\text{COOK}$ generates a strong base ($\text{KOH}$), so its solution is most basic.

9. The maximum prescribed concentration of copper in drinking water is: [JEE Main 2019]

(a) 5 ppm

(b) 0.5 ppm

(d) 3 ppm

Answer: (d)

Explanation: The maximum prescribed concentration of $\text{Cu}$ in drinking water is 3 ppm.

10. The correct decreasing order for acid strength is: [JEE Main 2019]

(a) $\text{NO}_2\text{CH}_2\text{COOH} > \text{NCCH}_2\text{COOH} > \text{FCH}_2\text{COOH} > \text{ClCH}_2\text{COOH}$

(b) $\text{FCH}_2\text{COOH} > \text{NCCH}_2\text{COOH} > \text{NO}_2\text{CH}_2\text{COOH} > \text{ClCH}_2\text{COOH}$

(d) $\text{CNCH}_2\text{COOH} > \text{O}_2\text{NCH}_2\text{COOH} > \text{FCH}_2\text{COOH} > \text{ClCH}_2\text{COOH}$

Answer: (a)

Explanation: Electron withdrawing group ($-\text{I}$ effect) in increasing order enhances acidic strength:

$$\text{NO}_2\text{CH}_2\text{COOH} > \text{NCCH}_2\text{COOH} > \text{FCH}_2\text{COOH} > \text{ClCH}_2\text{COOH}$$

11. 20 mL of $0.1\text{ M H}_2\text{SO}_4$ solution is added to 30 mL of $0.2\text{ M NH}_4\text{OH}$ solution. The pH of the resultant mixture is: [$\text{p}K_b \text{ of NH}_4\text{OH} = 4.7$] [JEE Main 2019]

(a) 9.4

(b) 5.0

(d) 5.2

Answer: (c)

Explanation: $20\text{ mL } 0.1\text{ M H}_2\text{SO}_4 \implies \eta_{\text{H}^+} = 4\text{ millimoles}$

$30\text{ mL } 0.2\text{ M NH}_4\text{OH} \implies \eta_{\text{NH}_4\text{OH}} = 6\text{ millimoles}$

$$\text{NH}_4\text{OH} + \text{H}^+ \longrightarrow \text{NH}_4^+ + \text{H}_2\text{O}$$ $$\text{Initial: } 6 \quad 4 \quad 0 \quad 0$$ $$\text{Final: } 2 \quad 0 \quad 4 \quad 4$$

Solution is a basic buffer:

$$\text{pOH} = \text{p}K_b + \log\left(\frac{[\text{NH}_4^+]}{[\text{NH}_4\text{OH}]}\right) = 4.7 + \log\left(\frac{4}{2}\right) = 4.7 + 0.3 = 5.0$$

$$\text{pH} = 14 - 5 = 9.0$$

12. The increasing basicity order of the following compounds is: [JEE Main 2019]

(A) $\text{CH}_3\text{CH}_2\text{NH}_2$

(B) $\text{(CH}_3\text{CH}_2\text{)}_2\text{NH}$

(D) $\text{Ph-NH-CH}_3$

(a) (D) < (C) < (A) < (B)

(b) (A) < (B) < (D) < (C)

(d) (D) < (C) < (B) < (A)

Answer: (a)

Explanation: In (D) $\text{Ph-NH-CH}_3$, lone pair is delocalised into benzene ring (least basic). In (C) $\text{(CH}_3\text{)}_3\text{N}$, steric hindrance lowers hydration energy in aqueous medium. Thus order is (D) < (C) < (A) < (B).

13. The pH of rain water is approximately: [JEE Main 2019]

(a) 6.5

(b) 7.5

(d) 7.0

Answer: (c)

Explanation: pH of unpolluted rain water is approximately 5.6 due to dissolved $\text{CO}_2$.

14. Consider the given plots for a reaction obeying Arrhenius equation ($0^\circ\text{C} < T < 300^\circ\text{C}$): ($k$ and $E_a$ are rate constant and activation energy, respectively) [JEE Main 2019]

Choose the correct option:

(a) Both I and II are wrong

(b) I is wrong but II is right

(d) I is right but II is wrong

Answer: (d)

Explanation: As $E_a$ increases, $k$ decreases exponentially ($k = A e^{-E_a/RT}$), so Plot I ($k$ vs $E_a$) is correct. For Plot II ($k$ vs $T$), $k$ increases with temperature $T$, so Plot II showing a decrease is wrong. Thus, I is right but II is wrong.

15. If a reaction follows the Arrhenius equation, the plot $\ln k$ vs $\frac{1}{RT}$ gives straight line with a gradient ($-y$) unit. The energy required to activate the reactant is: [JEE Main 2019]

(a) $y$ unit

(b) $-y$ unit

(d) $y/R$ unit

Answer: (a)

Explanation: $\ln k = \ln A - \frac{E_a}{RT}$. Slope of $\ln k$ vs $\frac{1}{RT}$ is $-E_a$. Given gradient $= -y \implies -E_a = -y \implies E_a = y\text{ unit}$.

16. $\text{NaH}$ is an example of: [JEE Main 2019]

(a) Electron-rich hydride

(b) Molecular hydride

(d) Metallic hydride

Answer: (c)

Explanation: $\text{NaH}$ is an ionic hydride, which is also known as a saline or salt-like hydride.

17. $\text{K}_2\text{HgI}_4$ is 40% ionised in aqueous solution. The value of its van't Hoff factor ($i$) is: [JEE Main 2019]

(a) 1.8

(b) 2.2

(d) 1.6

Answer: (a)

Explanation: $\text{K}_2[\text{HgI}_4] \longrightarrow 2\text{K}^+ + [\text{HgI}_4]^{2-} \implies n = 3$

$$i = 1 + \alpha(n - 1) = 1 + 0.4(3 - 1) = 1.8$$

18. The correct structure of histidine in a strongly acidic solution ($\text{pH} = 2$) is: [JEE Main 2019]

(a)

$$\begin{array}{c} \text{H}_3\overset{\oplus}{\text{N}}-\text{CH}-\text{COOH} \\ | \\ \text{CH}_2 \\ | \\ \text{C}=\text{CH} \\ /\quad\backslash \\ \text{HN}\quad \overset{\oplus}{\text{NH}} \\ \backslash\quad/ \\ \text{CH} \end{array}$$

(b)

$$\begin{array}{c} \text{H}_3\overset{\oplus}{\text{N}}-\text{CH}-\text{COO}^- \\ | \\ \text{CH}_2 \\ | \\ \text{C}=\text{CH} \\ /\quad\backslash \\ \text{HN}\quad \overset{\oplus}{\text{NH}} \\ \backslash\quad/ \\ \text{CH} \end{array}$$

(c)

$$\begin{array}{c} \text{H}_3\overset{\oplus}{\text{N}}-\text{CH}-\text{COOH} \\ | \\ \text{CH}_2 \\ | \\ \text{C}=\text{CH} \\ /\quad\backslash \\ \text{N}\quad \text{NH}_2^+ \\ \backslash\quad/ \\ \text{CH} \end{array}$$

(d)

$$\begin{array}{c} \text{H}_3\overset{\oplus}{\text{N}}-\text{CH}-\text{COO}^- \\ | \\ \text{CH}_2 \\ | \\ \text{C}=\text{CH} \\ /\quad\backslash \\ \text{N}\quad \text{NH}_2^+ \\ \backslash\quad/ \\ \text{CH} \end{array}$$

Answer: (a)

Explanation: At low $\text{pH} = 2$ (strongly acidic medium), all basic sites (amino group, imidazole ring nitrogen) are fully protonated and the carboxylate group remains as $-\text{COOH}$, giving a dicationic species.

Matrix Match/Column Match

19. Dilution processes of different aqueous solutions, with water, are given in Column I. The effects of dilution of the solutions on $[\text{H}^+]$ are given in Column II.

(Note: Degree of dissociation ($\alpha$) of weak acid and weak base is $\ll 1$; degree of hydrolysis of salt $\ll 1$; $[\text{H}^+]$ represents the concentration of $\text{H}^+$ ions) [JEE Adv. 2018 P-II]

Column I:

(A) (10 mL of 0.1 M $\text{NaOH}$ + 20 mL of 0.1 M acetic acid) diluted to 60 mL

(B) (20 mL of 0.1 M $\text{NaOH}$ + 20 mL of 0.1 M acetic acid) diluted to 80 mL

(D) 10 mL saturated solution of $\text{Ni(OH)}_2$ in equilibrium with excess solid $\text{Ni(OH)}_2$ is diluted to 20 mL (solid $\text{Ni(OH)}_2$ is still present after dilution).

Column II:

1. The value of $[\text{H}^+]$ does not change on dilution

2. The value of $[\text{H}^+]$ changes to half of its initial value on dilution

3. The value of $[\text{H}^+]$ changes to two times of its initial value on dilution

4. The value of $[\text{H}^+]$ changes to $\frac{1}{\sqrt{2}}$ times of its initial value on dilution.

5. The value of $[\text{H}^+]$ changes to $\sqrt{2}$ times of its initial value on dilution.

(a) A $\rightarrow$ 1; B $\rightarrow$ 5; C $\rightarrow$ 4; D $\rightarrow$ 1

(b) A $\rightarrow$ 2; B $\rightarrow$ 5; C $\rightarrow$ 4; D $\rightarrow$ 3

(d) A $\rightarrow$ 1; B $\rightarrow$ 5; C $\rightarrow$ 3; D $\rightarrow$ 2

Answer: (a)

Explanation:

(A) $[\text{CH}_3\text{COOH}]_{\text{old}} = \frac{20 \times 0.1 - 10 \times 0.1}{30} = \frac{1}{30}\text{ M}$, $[\text{CH}_3\text{COO}^-]_{\text{old}} = \frac{1}{30}\text{ M}$. Acidic buffer with $[\text{Salt}] = [\text{Acid}]$, so $[\text{H}^+] = K_a$ which does not change on dilution. $\implies \text{A} \rightarrow 1$.

(B) Salt of weak acid and strong base ($\text{CH}_3\text{COONa}$). $[\text{OH}^-] = \sqrt{\frac{K_w K_h}{c}}$. Dilution halves the concentration $c \longrightarrow c/2$, so $[\text{OH}^-]_{\text{new}} = \frac{[\text{OH}^-]_{\text{old}}}{\sqrt{2}}$, which means $[\text{H}^+]_{\text{new}} = \sqrt{2} [\text{H}^+]_{\text{old}}$. $\implies \text{B} \rightarrow 5$.

(C) Salt of strong acid and weak base ($\text{NH}_4\text{Cl}$). $[\text{H}^+] = \sqrt{\frac{K_w c}{K_b}}$. Dilution halves $c$, so $[\text{H}^+]_{\text{new}} = \frac{[\text{H}^+]_{\text{old}}}{\sqrt{2}}$. $\implies \text{C} \rightarrow 4$.

(D) For a saturated solution in equilibrium with solid solute, $[\text{OH}^-] = \sqrt[3]{2 K_{sp}}$ which is independent of solution volume. Hence $[\text{H}^+]$ remains constant. $\implies \text{D} \rightarrow 1$.

Paragraph

Paragraph for Questions Nos. 20 to 21

When 100 mL of 1.0 M $\text{HCl}$ was mixed with 100 mL of 1.0 M $\text{NaOH}$ in an insulated beaker at constant pressure, a temperature increase of $5.7^\circ\text{C}$ was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant ($-57.0\text{ kJ mol}^{-1}$), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid ($K_a = 2.0 \times 10^{-5}$) was mixed with 100 mL of 1.0 M $\text{NaOH}$ (under identical conditions to Expt. 1) where a temperature rise of $5.6^\circ\text{C}$ was measured. (Consider heat capacity of all solutions as $4.2\text{ J g}^{-1} \text{K}^{-1}$ and density of all solutions as $1.0\text{ g mL}^{-1}$)

20. Enthalpy of dissociation (in $\text{kJ mol}^{-1}$) of acetic acid obtained from Expt. 2 is: [JEE Adv. 2015 P-II]

(a) 1.0

(b) 10.0

(d) 51.4

Answer: (a)

Explanation: $n_{\text{HCl}} = 100 \times 1.0 = 100\text{ mmol} = 0.1\text{ mole}$.

Energy evolved due to neutralization $= 0.1 \times 57.0\text{ kJ} = 5.7\text{ kJ} = 5700\text{ J}$.

Energy used to increase temperature of solution $= 200\text{ g} \times 4.2\text{ J g}^{-1}\text{K}^{-1} \times 5.7\text{ K} = 4788\text{ J}$.

Energy absorbed by calorimeter $= 5700 - 4788 = 912\text{ J}$.

Calorimeter constant $C_{\text{cal}} = \frac{912}{5.7} = 160\text{ J/}^\circ\text{C}$.

In Expt. 2, energy evolved $= (200 \times 4.2 \times 5.6) + (160 \times 5.6) = 4704 + 896 = 5600\text{ J}$.

Energy used in dissociation of 0.1 mole $\text{CH}_3\text{COOH} = 5700 - 5600 = 100\text{ J} = 0.1\text{ kJ}$.

Enthalpy of dissociation $= \frac{0.1\text{ kJ}}{0.1\text{ mole}} = 1.0\text{ kJ/mole}$.

21. The pH of the solution after Expt. 2 is : [JEE Adv. 2015 P-II]

(a) 2.8

(b) 4.7

(d) 7.0

Answer: (b)

Explanation: $[\text{CH}_3\text{COOH}] = \frac{200 - 100}{200} = 0.5\text{ M}$

$[\text{CH}_3\text{COONa}] = \frac{100}{200} = 0.5\text{ M}$

$$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right) = -\log(2.0 \times 10^{-5}) + \log\left(\frac{0.5}{0.5}\right) = 4.7 + 0 = 4.7$$

Integer and Subjective

22. We have taken a saturated solution of $\text{AgBr}$. $K_{sp}$ of $\text{AgBr}$ is $12 \times 10^{-14}$. If $10^{-7}\text{ moles}$ of $\text{AgNO}_3$ are added to 1 litre of this solution find conductivity (specific conductance) of this solution in terms of $10^{-7}\text{ S m}^{-1}$ units. Given, $\lambda^\circ(\text{Ag}^+) = 6 \times 10^{-3}\text{ S m}^2\text{ mol}^{-1}$, $\lambda^\circ(\text{Br}^-) = 8 \times 10^{-3}\text{ S m}^2\text{ mol}^{-1}$, $\lambda^\circ(\text{NO}_3^-) = 7 \times 10^{-3}\text{ S m}^2\text{ mol}^{-1}$. [JEE 2006]

Answer: 55

Explanation: Solubility of $\text{AgBr}$ in presence of $10^{-7}\text{ M AgNO}_3$:

$$[\text{Ag}^+][\text{Br}^-] = (s + 10^{-7})s = 12 \times 10^{-14} \implies s = 3 \times 10^{-4}\text{ mol m}^{-3}$$

$$[\text{Br}^-] = 3 \times 10^{-4}\text{ mol m}^{-3}, \quad [\text{Ag}^+] = 4 \times 10^{-4}\text{ mol m}^{-3}, \quad [\text{NO}_3^-] = 10^{-4}\text{ mol m}^{-3}$$

$$\kappa_{\text{total}} = \kappa_{\text{Br}^-} + \kappa_{\text{Ag}^+} + \kappa_{\text{NO}_3^-}$$

$$\kappa = (3 \times 10^{-4} \times 8 \times 10^{-3}) + (4 \times 10^{-4} \times 6 \times 10^{-3}) + (10^{-4} \times 7 \times 10^{-3}) = (24 + 24 + 7) \times 10^{-7} = 55 \times 10^{-7}\text{ S m}^{-1}$$

23. The dissociation constant of a substituted benzoic acid at $25^\circ\text{C}$ is $1.0 \times 10^{-4}$. The pH of a 0.01M solution of its sodium salt is: [JEE 2009 P-II]

Answer: 8

Explanation: $K_a = 1.0 \times 10^{-4}$, $C = 0.01\text{ M}$

$$\text{C}_6\text{H}_5\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{COOH} + \text{OH}^-$$ $$0.01(1-h) \qquad 0.01h \qquad 0.01h$$

$$K_h = \frac{K_w}{K_a} = \frac{10^{-14}}{10^{-4}} = 10^{-10} = \frac{C h^2}{1-h} \approx 0.01 h^2$$

$$h^2 = 10^{-8} \implies h = 10^{-4}$$

$$[\text{OH}^-] = C h = 0.01 \times 10^{-4} = 10^{-6}\text{ M}$$

$$[\text{H}^+] = \frac{10^{-14}}{10^{-6}} = 10^{-8}\text{ M} \implies \text{pH} = 8$$

24. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is: $\text{KCN}$, $\text{K}_2\text{SO}_4$, $(\text{NH}_4)_2\text{C}_2\text{O}_4$, $\text{NaCl}$, $\text{Zn(NO}_3)_2$, $\text{FeCl}_3$, $\text{K}_2\text{CO}_3$, $\text{NH}_4\text{NO}_3$, $\text{LiCN}$ [JEE 2010 P-I]

Answer: 3

Explanation: Compounds forming basic aqueous solutions (salts of weak acids and strong bases) will turn red litmus blue. Among the given compounds, $\text{KCN}$, $\text{K}_2\text{CO}_3$, and $\text{LiCN}$ are basic in nature. So, answer is 3.

25. The solubility of a salt of weak acid ($\text{AB}$) at pH 3 is $Y \times 10^{-3}\text{ mol L}^{-1}$. The value of Y is ______. (Given that the value of solubility product of $\text{AB}$ ($K_{sp}$) $= 2 \times 10^{-10}$ and the value of ionization constant of $\text{HB}$ ($K_a$) $= 1 \times 10^{-8}$) [JEE Adv. 2018 P-I]

Answer: 4.47

Explanation: $\text{AB}(s) \rightleftharpoons \text{A}^+(aq) + \text{B}^-(aq)$ (solubility $= s$)

$\text{B}^- + \text{H}^+ \rightleftharpoons \text{HB}$

$$K_a = \frac{[\text{H}^+][\text{B}^-]}{[\text{HB}]} = 1 \times 10^{-8}$$

At $\text{pH} = 3 \implies [\text{H}^+] = 10^{-3}\text{ M}$

$$\frac{10^{-3} [\text{B}^-]}{[\text{HB}]} = 10^{-8} \implies [\text{HB}] = 10^5 [\text{B}^-]$$

Also, $[\text{A}^+] = s$, $[\text{B}^-] + [\text{HB}] = s \implies [\text{B}^-] + 10^5 [\text{B}^-] = s \implies [\text{B}^-] \approx \frac{s}{10^5}$$

$$K_{sp} = [\text{A}^+][\text{B}^-] = s \times \frac{s}{10^5} = 2 \times 10^{-10}$$

$$s^2 = 2 \times 10^{-5} = 20 \times 10^{-6}$$

$$s = \sqrt{20} \times 10^{-3} \approx 4.47 \times 10^{-3}\text{ mol L}^{-1}$$

$$\text{So, } Y = 4.47$$