Sulphurous acid ( H_2SO_3 ) has Ka1 = 1.7 \times 10 - 2 and Ka2 = 6.4 \times 10 - 8. The pH of 0.588 M H_2SO_3 is \underline\hspace2cm . (Round off to the Nearest Integer).

Ionic Equilibrium PYQ — JEE Main 2021 · 16 Mar · Shift 2 · Q22 with Solution

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Correct answer: 1

Given data

Sulphurous acid is a diprotic acid:

ightleftharpoons H^+ + HSO_3^- , , \quad K_{a1}=1.7\times 10^{-2}$$

ightleftharpoons H^+ + SO_3^{2-} \, , \quad K_{a2}=6.4\times 10^{-8}$$ Initial concentration: $$C=0.588\,\text{M}$$ 2. **Check whether second dissociation matters** Since $$K_{a2}=6.4\times 10^{-8} \ll K_{a1}=1.7\times 10^{-2},$$ the second dissociation contributes negligibly to $[H^+]$ compared to the first. So we first treat $H_2SO_3$ as a weak monoprotic acid with concentration $0.588\,\text{M}$. 3. **First dissociation equilibrium** Let $x=[H^+]$ produced from the first step. Then at equilibrium: $$[H_2SO_3]=0.588-x, \quad [H^+]=x, \quad [HSO_3^-]=x$$ Apply $K_{a1}$: $$K_{a1}=\frac{x^2}{0.588-x}=1.7\times 10^{-2}$$ So, $$x^2=0.017(0.588-x)$$ $$x^2=0.009996-0.017x$$ $$x^2+0.017x-0.009996=0$$ 4. **Solve the quadratic** $$x=\frac{-0.017+\sqrt{(0.017)^2+4(0.009996)}}{2}$$ $$x=\frac{-0.017+\sqrt{0.000289+0.039984}}{2}$$ $$x=\frac{-0.017+\sqrt{0.040273}}{2}$$ $$\sqrt{0.040273}\approx 0.2007$$ Hence, $$x\approx \frac{-0.017+0.2007}{2}=\frac{0.1837}{2}=0.09185$$ Therefore, $$[H^+]\approx 0.09185\,\text{M}$$ 5. **Check contribution from second dissociation** For the second step, $$K_{a2}=\frac{[H^+][SO_3^{2-}]}{[HSO_3^-]}$$ Since after first dissociation, $[H^+]\approx [HSO_3^-]\approx 0.09185$, let additional $H^+$ from second step be $y$. Then approximately, $$K_{a2}\approx y$$ because $$K_{a2}\approx \frac{(0.09185)y}{0.09185}=y$$ So, $$y\approx 6.4\times 10^{-8}$$ This is negligible compared to $0.09185$. Thus total $$[H^+]\approx 0.09185\,\text{M}$$ 6. **Calculate pH** $$\text{pH}=-\log(0.09185)$$ $$\text{pH}\approx 1.04$$ 7. **Nearest integer** $$\boxed{1}$$

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View written solutionFree

Correct answer: 1

Given data

Sulphurous acid is a diprotic acid:

ightleftharpoons H^+ + HSO_3^- , , \quad K_{a1}=1.7\times 10^{-2}$$

ightleftharpoons H^+ + SO_3^{2-} \, , \quad K_{a2}=6.4\times 10^{-8}$$ Initial concentration: $$C=0.588\,\text{M}$$ 2. **Check whether second dissociation matters** Since $$K_{a2}=6.4\times 10^{-8} \ll K_{a1}=1.7\times 10^{-2},$$ the second dissociation contributes negligibly to $[H^+]$ compared to the first. So we first treat $H_2SO_3$ as a weak monoprotic acid with concentration $0.588\,\text{M}$. 3. **First dissociation equilibrium** Let $x=[H^+]$ produced from the first step. Then at equilibrium: $$[H_2SO_3]=0.588-x, \quad [H^+]=x, \quad [HSO_3^-]=x$$ Apply $K_{a1}$: $$K_{a1}=\frac{x^2}{0.588-x}=1.7\times 10^{-2}$$ So, $$x^2=0.017(0.588-x)$$ $$x^2=0.009996-0.017x$$ $$x^2+0.017x-0.009996=0$$ 4. **Solve the quadratic** $$x=\frac{-0.017+\sqrt{(0.017)^2+4(0.009996)}}{2}$$ $$x=\frac{-0.017+\sqrt{0.000289+0.039984}}{2}$$ $$x=\frac{-0.017+\sqrt{0.040273}}{2}$$ $$\sqrt{0.040273}\approx 0.2007$$ Hence, $$x\approx \frac{-0.017+0.2007}{2}=\frac{0.1837}{2}=0.09185$$ Therefore, $$[H^+]\approx 0.09185\,\text{M}$$ 5. **Check contribution from second dissociation** For the second step, $$K_{a2}=\frac{[H^+][SO_3^{2-}]}{[HSO_3^-]}$$ Since after first dissociation, $[H^+]\approx [HSO_3^-]\approx 0.09185$, let additional $H^+$ from second step be $y$. Then approximately, $$K_{a2}\approx y$$ because $$K_{a2}\approx \frac{(0.09185)y}{0.09185}=y$$ So, $$y\approx 6.4\times 10^{-8}$$ This is negligible compared to $0.09185$. Thus total $$[H^+]\approx 0.09185\,\text{M}$$ 6. **Calculate pH** $$\text{pH}=-\log(0.09185)$$ $$\text{pH}\approx 1.04$$ 7. **Nearest integer** $$\boxed{1}$$

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