JEE Advanced 2025 Paper 1 · Q09 · Ionic Equilibrium & pH
At 25 °C, the $[H^+]$ in a $1.00 \times 10^{-3}$ M aqueous solution of a weak monobasic acid with $K_a = 4.00 \times 10^{-11}$ is $X \times 10^{-7}$ M. Find X. Use $K_w = 1.00 \times 10^{-14}$.
Reveal answer + step-by-step solution
Correct answer:2.24
Solution
Both the acid and water dissociation contribute to $[H^+]$; the acid is so weak that one cannot ignore $K_w$. Charge balance: $[H^+] = [A^-] + [OH^-]$; Mass balance: $[HA] + [A^-] = C_0 = 10^{-3}$. Since $K_a$ is tiny, $[A^-] \ll C_0$, so $[HA] \approx C_0$. Then $[A^-] = K_a [HA] / [H^+] \approx K_a C_0 / [H^+]$. Also $[OH^-] = K_w/[H^+]$. $[H^+] = K_a C_0 / [H^+] + K_w / [H^+]$ ⟹ $[H^+]^2 = K_a C_0 + K_w = (4.00 \times 10^{-11})(10^{-3}) + 10^{-14} = 4.00 \times 10^{-14} + 1.00 \times 10^{-14} = 5.00 \times 10^{-14}$. $[H^+] = \sqrt{5} \times 10^{-7} \approx 2.236 \times 10^{-7}$ M. So $X \approx 2.24$ (within official band $[2.2, 2.3]$).
Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →