50 mL of 0.02 M NaHSO_4 is mixed with 50 m | vedclass

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(B) The total volume of the mixture is $100 \ mL$. The concentration of $HSO_4^-$ is $0.01 \ M$ and $SO_4^{2-}$ is $0.01 \ M$ after mixing.$HSO_4^- \r

Vedclass · Accepted Answer

$2 - \log (\sqrt{2} - 1)$

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(B) The total volume of the mixture is $100 \ mL$. The concentration of $HSO_4^-$ is $0.01 \ M$ and $SO_4^{2-}$ is $0.01 \ M$ after mixing.$HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}$Initial concentrations: $[HSO_4^-] = 0.01 \ M$,$[SO_4^{2-}] = 0.01 \ M$,$[H^+] = 0 \ M$.At equilibrium: $[HSO_4^-] = 0.01(1-\alpha)$,$[H^+] = 0.01\alpha$,$[SO_4^{2-}] = 0.01 + 0.01\alpha$.Given $pK_{a2} = 2$,so $K_a = 10^{-2}$.$10^{-2} = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]} = \frac{(0.01\alpha)(0.01 + 0.01\alpha)}{0.01(1-\alpha)} = \frac{0.01\alpha(1+\alpha)}{1-\alpha}$.$1 - \alpha = \alpha + \alpha^2 \Rightarrow \alpha^2 + 2\alpha - 1 = 0$.Solving for $\alpha$: $\alpha = \frac{-2 + \sqrt{4 - 4(1)(-1)}}{2} = \sqrt{2} - 1$.$[H^+] = 0.01\alpha = 10^{-2}(\sqrt{2} - 1)$.$pH = -\log[H^+] = -\log(10^{-2}(\sqrt{2} - 1)) = 2 - \log(\sqrt{2} - 1)$.

$50 \ mL$ of $0.02 \ M$ $NaHSO_4$ is mixed with $50 \ mL$ of $0.02 \ M$ $Na_2SO_4$. Calculate the $pH$ of the resulting solution. $[pK_{a2} (H_2SO_4) = 2]$

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