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  • A hydrogen electrode placed in a solution containing CH3COOK and CH3COOH in the ratio of a : b and b : a has electrode potential values of -1.59 and +1 V, respectively. Calcul

A hydrogen electrode placed in a solution containing CH3COOK and CH3COOH in the ratio of a : b and b : a has electrode potential values of -1.59 and +1 V, respectively. Calculate pKa of CH3COOH.

Option: 1

5


Option: 2

5.9


Option: 3

10.8


Option: 4

10


Answers (1)

best_answer

Mixture of CH3COOK and CH3COOH forms acidic buffer solution whose pH is:
\mathrm{pH}_{\text { {acidic buffer}}}=\mathrm{p} \mathrm K_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\text { Acid }]}
\therefore \: \mathrm E_1=-0.059 \mathrm{pH}_{1}=-0.059\left[\mathrm{p} \mathrm K_{\mathrm{a}}+\log \frac{a}{b}\right]\: \: \: \: \: \: ...(\mathrm i)
   \mathrm E_{2}=-0.059\: \mathrm{pH}_{2}=-0.059\left[\mathrm{p}\mathrm K_{\mathrm{a}}+\log \frac{b}{a}\right]\: \: \: \: \: \: ...(\mathrm {ii})
\text { Adding Eqs.(i) and (ii), we get }
\mathrm{E_{1}+E_{2}}=-0.059\left[\mathrm{p K}_{\mathrm{a}}+\log \frac{a}{b}\right]-0.059\left[\mathrm{p K}_{\mathrm{a}}+\log \frac{b}{a}\right]
                  =-2 \times 0.059 \mathrm{\: p K}_{\mathrm{a}}
\therefore \mathrm{p K}_{\mathrm{a}}=\frac{-\left(E_{1}+E_{2}\right)}{2 \times 0.059}=\frac{-(-1.59+1.0)}{2 \times 0.059}=\frac{0.59}{2 \times 0.059}=5

\therefore \mathrm{p K}_{\mathrm{a}}=5

 

Posted by

manish painkra

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