For , and . The molarity of solution of pH 2.0 is ''. The value of '1000000' is — Ionic Equilibrium Chemistry Question
Question
For $\text{H}_2\text{SO}_4$, $K_{a1} = \text{infinite}$ and $K_{a2} = 1.2 \times 10^{-2}$. The molarity of $\text{H}_2\text{SO}_4$ solution of pH 2.0 is '$x\text{ M}$'. The value of '1000000$x$' is
💡 Solution & Explanation
Given $\text{pH} = 2.0$, the total $[\text{H}^+] = 10^{-2}\text{ M} = 0.01\text{ M}$. Since $K_{a1}$ is infinite, the first dissociation gives $[\text{H}^+] = x$ and $[\text{HSO}_4^-] = x$. Let the second dissociation produce $y$ moles of $\text{H}^+$, so total $[\text{H}^+] = x + y = 0.01$. At equilibrium, $[\text{HSO}_4^-] = x - y$ and $[\text{SO}_4^{2-}] = y$. From $K_{a2}$, we have $1.2 \times 10^{-2} = \frac{[\text{H}^+][\text{SO}_4^{2-}]}{[\text{HSO}_4^-]} = \frac{0.01 \times y}{x - y}$. Thus, $1.2(x - y) = y \implies 1.2x = 2.2y \implies y = \frac{1.2}{2.2}x$. Substituting into $x + y = 0.01$, we get $x + \frac{1.2}{2.2}x = 0.01 \implies \frac{3.4}{2.2}x = 0.01 \implies x = \frac{0.022}{3.4} \approx 0.0064705\text{ M}$. $1000000x = 6470.5$, which rounds to 6470.