50 ml of 0.02 M NaHSO_4 is mixed with 50 ml o | vedclass

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Question

$\mathop {\mathop {HSO_4^ - }\limits_{0.01} }\limits_{0.01(1 - \alpha )} \xrightarrow{{{{10}^{ - 2}}}}\mathop {\mathop {{H^ + }}\limits_{} }\limits_{0

Vedclass · Accepted Answer

$2 - \,\log \,\left( {\sqrt 2 } \right) - 1$

Vedclass · Answer

$\mathop {\mathop {HSO_4^ - }\limits_{0.01} }\limits_{0.01(1 - \alpha )} \xrightarrow{{{{10}^{ - 2}}}}\mathop {\mathop {{H^ + }}\limits_{} }\limits_{0.01\alpha }  + \mathop {\mathop {SO_4^{ -  - }}\limits_{0.01} }\limits_{0.01 + 0.01\alpha } $

$10^{-2}=\frac{0.01 \alpha 	imes 0.01(1+\alpha)}{0.01(1-\alpha)}$

$(1-\alpha)=\alpha^{2}+\alpha$

$\alpha^{2}+2 \alpha-1=0$

$\alpha=\frac{-2+\sqrt{4+4}}{2}$

$=(-1+\sqrt{2})$

$\left(\mathrm{H}^{+}ight)=(\sqrt{2}-1) 	imes 10^{-2}$

$\mathrm{pH}=2-\log (\sqrt{2}-1)$

$50\ ml$ of $0.02\ M$ $NaHSO_4$ is mixed with $50$ $ml$ of $0.02\ M\ Na_2SO_4$. Calculate $pH$ of the resulting solution.$[pKa_2 (H_2SO_4) = 2]$

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