Q.The ionization constant of chloroacetic acid is 1.35 × 10⁻³. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?

Showing the 12 most recent of 16 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The dissociation constants of $H_2A$ are $K_{a1} = 6 \times 10^{-2}$ and $K_{a2} = 6 \times 10^{-5}$ respectively. At equilibrium, $[A^{2-}] = [H_2A]$. What is the approximate concentration of $H^+$ at equilibrium? (A) $1.9 \times 10^{-3}$ (B) $2 \times 10^{-4}$ (C) $1.9 \times 10^{-5}$ (D) $1.9 \times 10^{-2}$

   ›Reveal solutionSolution

   A classic diprotic-acid trick: combining $K_{a1}$ and $K_{a2}$ eliminates the intermediate species $HA^-$ when $[A^{2-}]=[H_2A]$, letting $[H^+]$ be found directly from $\sqrt{K_{a1}K_{a2}}$.

   Concept and Intuition

   For a diprotic acid $H_2A$, the two successive dissociations are:

   $H_2A \rightleftharpoons H^+ + HA^- \qquad K_{a1} = \frac{[H^+][HA^-]}{[H_2A]}$

   $HA^- \rightleftharpoons H^+ + A^{2-} \qquad K_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]}$

   Multiplying these two expressions together makes the intermediate concentration $[HA^-]$ cancel out entirely:

   $K_{a1}K_{a2} = \frac{[H^+]^2[A^{2-}]}{[H_2A]}$

   This is a very useful identity whenever you're told something about the relationship between $[A^{2-}]$ and $[H_2A]$ directly, bypassing the need to know $[HA^-]$ at all.

   Step-by-Step Solution

   1. Write $K_{a1}K_{a2} = \dfrac{[H^+]^2[A^{2-}]}{[H_2A]}$.
   2. Given $[A^{2-}] = [H_2A]$, the fraction $\dfrac{[A^{2-}]}{[H_2A]} = 1$.
   3. So $K_{a1}K_{a2} = [H^+]^2$.
   4. Substitute: $[H^+]^2 = (6\times10^{-2})(6\times10^{-5}) = 36\times10^{-7} = 3.6\times10^{-6}$. …

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2. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.What is the $pH$ of 0.05 M $HCN$ solution? ($K_a = 5\times10^{-10}$; $\log 5 = 0.7$) (A) 5.3 (B) 6.3 (C) 4.3 (D) 4.7

   ›Reveal solutionSolution

   Weak-acid pH formula $[H^+]=\sqrt{K_aC}$ gives $[H^+]=5\times10^{-6}$ M and $pH=5.3$.

   Concept and Intuition

   For a weak monoprotic acid HA with small degree of ionisation, the equilibrium $[H^+]\approx[A^-]$ and $[HA]\approx C$ (initial concentration), so $K_a = \dfrac{[H^+]^2}{C}$, giving the shortcut $[H^+]=\sqrt{K_aC}$.

   Step-by-Step Solution

   1. $[H^+] = \sqrt{K_a \times C} = \sqrt{5\times10^{-10}\times 0.05}$.
   2. $5\times10^{-10}\times0.05 = 2.5\times10^{-11}$.
   3. $\sqrt{2.5\times10^{-11}} = \sqrt{25\times10^{-12}} = 5\times10^{-6}$ M. …

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3. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.The pH of a 0.1 M solution of a weak monobasic organic acid is 4.0. What is the dissociation constant of the acid? (A) $1.0\times10^{-8}$ (B) $1.0\times10^{-7}$ (C) $1.0\times10^{-6}$ (D) $1.0\times10^{-5}$

   ›Reveal solutionSolution

   Standard weak-acid dissociation calculation from pH: $K_a = [H^+]^2/C$. Answer: $1.0\times10^{-7}$.

   Concept and Intuition

   For a weak monobasic acid HA dissociating as $HA \rightleftharpoons H^+ + A^-$, if $\alpha$ is the small degree of dissociation, $[H^+]=C\alpha$ and $K_a = \dfrac{C\alpha^2}{1-\alpha}\approx C\alpha^2$ (since $\alpha\ll1$ for a weak acid). This simplifies to $K_a\approx[H^+]^2/C$.

   Step-by-Step Solution

   1. $pH=4.0 \Rightarrow [H^+]=10^{-4}$ M.
   2. Concentration $C=0.1$ M.
   3. $K_a = \dfrac{[H^+]^2}{C} = \dfrac{(10^{-4})^2}{0.1} = \dfrac{10^{-8}}{10^{-1}} = 10^{-7}$.
   4. Check: $\alpha = [H^+]/C = 10^{-4}/0.1 = 10^{-3}$, which is indeed $\ll 1$, validating the weak-acid approximation.

   Common Mistakes …

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4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.At $25°C$, the percentage of ionization of $x$ M acetic acid is 4.242. What is the pH of the acetic acid solution? ($\log 4.242 = 0.6275$); ($\log 0.04242 = -1.372$) ($K_a = 1.8 \times 10^{-5}$) (A) 3.37 (B) 1.70 (C) 1.37 (D) 2.37

   ›Reveal solutionSolution

   This tests Ostwald's dilution law for a weak acid: using $K_a = C\alpha^2$ to back out the concentration from the given percentage ionisation, then computing $[H^+]=C\alpha$ and finally the pH, gives 3.37.

   Concept and Intuition

   For a weak monoprotic acid HA that ionises to a small fraction $\alpha$ (degree of dissociation), the equilibrium constant is $K_a = \dfrac{C\alpha^2}{1-\alpha} \approx C\alpha^2$ for small $\alpha$. Given $K_a$ and $\alpha$, we can solve backward for the concentration $C$, and from there for $[H^+] = C\alpha$ and hence pH. The provided log values are the exact hint that this problem wants a logarithmic (not linear) route to the answer.

   Step-by-Step Solution

   1. Convert percentage ionisation to a fraction: $\alpha = 4.242\% = 0.04242$.
   2. Apply Ostwald's dilution law: $K_a = C\alpha^2 \Rightarrow C = \dfrac{K_a}{\alpha^2}$.
   3. Take logs: $\log C = \log K_a - 2\log\alpha$.
   4. $\log K_a = \log(1.8\times10^{-5}) = \log1.8 - 5 \approx 0.2553 - 5 = -4.7447$.
   5. $\log\alpha = \log(0.04242) = -1.372$ (given).
   6. $\log C = -4.7447 - 2(-1.372) = -4.7447 + 2.744 = -2.0007 \approx -2.00$, so $C \approx 1.0\times10^{-2}$ M. …

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5. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.At 25 °C, the percentage of ionization of 'x' M acetic acid is 4.242. What is the value of x ? ($K_a = 1.8\times10^{-5}$). (A) 0.05 (B) 0.04 (C) 0.02 (D) 0.01

   ›Reveal solutionSolution

   Tests the weak-acid degree-of-ionization formula $\alpha = \sqrt{K_a/C}$; solving for concentration gives $x = 0.01$ M.

   Concept and Intuition

   For a weak monoprotic acid $HA \rightleftharpoons H^+ + A^-$ with initial concentration $C$ and degree of dissociation $\alpha$ (small, so $1-\alpha \approx 1$), the equilibrium constant is approximately $K_a \approx C\alpha^2$, giving $\alpha \approx \sqrt{K_a/C}$. Rearranging lets us find the concentration from a known percentage ionization.

   Step-by-Step Solution

   1. Convert percentage ionization to a fraction: $\alpha = \dfrac{4.242}{100} = 0.04242$.
   2. Use the approximate relation: $K_a = C\alpha^2 \Rightarrow x = \dfrac{K_a}{\alpha^2}$.
   3. Compute $\alpha^2 = (0.04242)^2 \approx 1.8\times10^{-3}$. …

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6. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.At $25\,^\circ\mathrm{C}$, $K_a$ of acetic acid is $1.8\times10^{-5}$. What is the percentage of ionization of 0.02 M acetic acid at this temperature? (A) $3$ (B) $4.242$ (C) $5$ (D) $1.414$

   ›Reveal solutionSolution

   Ostwald's dilution law gives the degree of ionisation of a weak acid as $\sqrt{K_a/C}$; here it works out to $3\%$.

   Concept and Intuition

   For a weak monoprotic acid $\mathrm{HA}\rightleftharpoons H^++A^-$ with initial concentration $C$ and degree of dissociation $\alpha$ (small, so $1-\alpha\approx1$), $K_a \approx C\alpha^2$, giving $\alpha = \sqrt{K_a/C}$ (Ostwald's dilution law).

   Step-by-Step Solution

   1. $K_a = 1.8\times10^{-5}$, $C=0.02\,\mathrm{M}$.
   2. $\alpha = \sqrt{\dfrac{1.8\times10^{-5}}{0.02}} = \sqrt{9\times10^{-4}} = 3\times10^{-2} = 0.03$. …

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7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.The percentage of ionization of 1 L of $x$ M acetic acid is 4.242 and is called solution "A". The percentage of ionization of 1 L of $y$ M acetic acid is 3 and is called solution "B". Solution "A" is mixed with solution "B". What is the concentration of acetic acid in the resultant solution? ($K_a$ of acetic acid $= 1.8\times10^{-5}$) (A) 0.05 M (B) 0.015 M (C) 0.02 M (D) 0.15 M

   ›Reveal solutionSolution

   Back out each solution's initial concentration from its % ionization via $K_a=C\alpha^2$, then average the two moles over the combined 2 L volume — giving 0.015 M.

   Concept and Intuition

   For a weak acid at low degree of dissociation $\alpha$, the Ostwald dilution law simplifies to $K_a \approx C\alpha^2$ (since $1-\alpha \approx 1$). Given $K_a$ and $\alpha$ for each solution, we can solve backward for each solution's original concentration $C$. When two solutions of a weak acid (not just its ions) are mixed, the total moles of acetic acid (dissociated + undissociated) simply add, and the new concentration is total moles over total volume.

   Step-by-Step Solution

   1. Solution A: $\alpha_A = 4.242\% = 0.04242$. Using $K_a = C\alpha^2$: $C_A = \dfrac{1.8\times10^{-5}}{(0.04242)^2} = \dfrac{1.8\times10^{-5}}{1.8\times10^{-3}} = 0.01\ \mathrm{M}$.
   2. Solution B: $\alpha_B = 3\% = 0.03$. $C_B = \dfrac{1.8\times10^{-5}}{(0.03)^2} = \dfrac{1.8\times10^{-5}}{9\times10^{-4}} = 0.02\ \mathrm{M}$.
   3. Moles of acetic acid in 1 L of A = 0.01 mol; moles in 1 L of B = 0.02 mol. …

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8. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
   Q.The dissociation constants of $\mathrm{H_2A}$ are $K_{a1} = 6 \times 10^{-2}$; $K_{a2} = 6 \times 10^{-5}$. The pH of 0.011 M $\mathrm{H_2A}$ solution is 2.0. What is the value of $\frac{[A^{2-}]}{[H_2A]}$? (A) 0.036 (B) 0.36 (C) 3.6 (D) $36 \times 10^{-5}$

   ›Reveal solutionSolution

   Multiplying the two stepwise dissociation constants gives the overall $\mathrm{H_2A \rightleftharpoons 2H^+ + A^{2-}}$ equilibrium constant; dividing by $[H^+]^2$ (from the given pH) yields $[A^{2-}]/[H_2A] = 0.036$.

   Concept and Intuition

   For a diprotic acid, the two stepwise equilibria are:

   $\mathrm{H_2A \rightleftharpoons H^+ + HA^-}, \quad K_{a1} = \frac{[H^+][HA^-]}{[H_2A]}$

   $\mathrm{HA^- \rightleftharpoons H^+ + A^{2-}}, \quad K_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]}$

   Multiplying these two equilibrium expressions together, the $[HA^-]$ terms cancel, giving the overall two-proton dissociation:

   $K_{a1}K_{a2} = \frac{[H^+]^2[A^{2-}]}{[H_2A]}$

   Step-by-Step Solution

   1. $K_{a1}K_{a2} = (6\times10^{-2})(6\times10^{-5}) = 36\times10^{-7} = 3.6\times10^{-6}$.
   2. pH = 2.0 $\Rightarrow$ $[H^+] = 10^{-2}$ M, so $[H^+]^2 = 10^{-4}$. …

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9. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.At 298 K, the ionization constant of $\mathrm{CN^-}$ is $2.08 \times 10^{-6}$. What is the ionization constant of its conjugate acid? ($K_w = 10^{-14}$) (A) $2.08 \times 10^{8}$ (B) $4.8 \times 10^{-8}$ (C) $4.8 \times 10^{-9}$ (D) $2.08 \times 10^{-8}$

   ›Reveal solutionSolution

   For a conjugate acid-base pair, $K_aK_b=K_w$; using the given $K_b$ of $\mathrm{CN^-}$ gives $K_a(\mathrm{HCN})\approx4.8\times10^{-9}$.

   Concept and Intuition

   $\mathrm{CN^-}$ is the conjugate base of the weak acid HCN. Whenever a conjugate acid-base pair is involved, their ionization constants are linked through the autoionization constant of water: $K_a\times K_b = K_w$. This lets you get one from the other without doing a fresh equilibrium calculation.

   Step-by-Step Solution

   1. Given: ionization constant of $\mathrm{CN^-}$ (as a base, hydrolyzing water) $K_b = 2.08\times10^{-6}$.
   2. Relation: $K_a(\text{conjugate acid}) \times K_b(\text{base}) = K_w$. …

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10. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.At $27^0$C, the degree of dissociation of weak acid (HA) in its 0.5M aqueous solution is 1%. Its $K_a$ value is approximately (A) $5 \times 10^{-4}$ (B) $5 \times 10^{-5}$ (C) $5 \times 10^{-6}$ (D) $5 \times 10^{-8}$

    ›Reveal solutionSolution

    Standard weak-acid dissociation approximation $K_a \approx c\alpha^2$ gives $K_a = 5\times10^{-5}$ — option (B).

    Concept and Intuition

    For a weak acid HA with small degree of dissociation $\alpha$, the equilibrium expression $K_a = \dfrac{c\alpha^2}{1-\alpha}$ simplifies to $K_a \approx c\alpha^2$ because $\alpha \ll 1$ makes $(1-\alpha) \approx 1$. This is the standard shortcut used whenever $\alpha$ is a percent-level quantity like 1%.

    Step-by-Step Solution

    1. Given: $c = 0.5$ M, $\alpha = 1\% = 0.01$.
    2. $K_a = \dfrac{c\alpha^2}{1-\alpha} \approx c\alpha^2$ (valid since $\alpha = 0.01 \ll 1$).
    3. $K_a = 0.5 \times (0.01)^2 = 0.5 \times 0.0001 = 5\times10^{-5}$.

    Common Mistakes …

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11. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.At 27°C, the degree of dissociation of HA (weak acid) in 0.5 M of its solution is 1%. The concentrations of $\mathrm{H_3O^+}$, $\mathrm{A^-}$ and HA at equilibrium (in $\mathrm{mol\ L^{-1}}$) are respectively (A) 0.005, 0.005, 0.495 (B) 0.05, 0.05, 0.45 (C) 0.01, 0.01, 0.49 (D) 0.005, 0.495, 0.005

    ›Reveal solutionSolution

    Using the degree-of-dissociation ($\alpha$) formulas for a weak acid at equilibrium gives $[\mathrm{H_3O^+}]=[\mathrm{A^-}]=0.005\ \mathrm{M}$ and $[\mathrm{HA}]=0.495\ \mathrm{M}$.

    Concept and Intuition

    For a weak monoprotic acid $\mathrm{HA} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{A^-}$ starting at concentration $C$, if $\alpha$ is the fraction that dissociates at equilibrium, then the amount dissociated is $C\alpha$ (this becomes both $[\mathrm{H_3O^+}]$ and $[\mathrm{A^-}]$ in a 1:1 stoichiometry), and the amount remaining undissociated is $C(1-\alpha)$.

    Step-by-Step Solution

    1. Given: $C = 0.5\ \mathrm{M}$, $\alpha = 1\% = 0.01$.
    2. $[\mathrm{H_3O^+}] = C\alpha = 0.5 \times 0.01 = 0.005\ \mathrm{M}$.
    3. $[\mathrm{A^-}] = C\alpha = 0.005\ \mathrm{M}$ (same as H$_3$O$^+$ since 1 mole of each is produced per mole of HA dissociated). …

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12. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
    Q.A, B and C are weak acids. Their dissociation constants ($K_a(A)$, $K_a(B)$, $K_a(C)$) are $3.5\times10^4$, $1\times10^{-5}$ and $5\times10^{-10}$ respectively. The pH of 1L of 0.01 M each of these solutions follow the order (A) $C > B > A$ (B) $A > B > C$ (C) $B > A > C$ (D) $C > A > B$

    ›Reveal solutionSolution

    A larger dissociation constant $K_a$ means a stronger weak acid and hence a lower pH; ranking the given $K_a$ values in decreasing order (A > B > C) gives the pH order $C>B>A$.

    Concept and Intuition

    For a weak acid, $[H^+]=\sqrt{K_aC}$ at equal concentration $C$. Since $[H^+]$ increases with $K_a$, pH (which is inversely related to $[H^+]$) decreases as $K_a$ increases. So the acid with the largest $K_a$ has the smallest pH, and vice versa.

    Step-by-Step Solution

    1. List the given values: $K_a(A)=3.5\times10^{-4}$, $K_a(B)=1\times10^{-5}$, $K_a(C)=5\times10^{-10}$.
    2. Rank by acid strength (largest $K_a$ = strongest acid): $A > B > C$.
    3. Since stronger acids dissociate more, giving higher $[H^+]$ and hence lower pH, the pH ranking is the exact reverse of the acid-strength ranking. …

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