The dissociation constant of acetic acid is x×10-5. When 25 mL of 0.2 M CH3COONa solution is mixed with 25 mL of 0.02 M CH3COOH solution the pH of the resultant solution is found to be equal to 5. The value of x is______

Buffer soln :pH=pKa+log⁡[ salt ][ acid ]5=pKa+log⁡25×0.225×0.02⇒pKa=5−1=4⇒Ka = 10-4 = 10 x 10-5= 10

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The dissociation constant of acetic acid is $x \times 10^{- 5} .$ When 25 mL of 0.2 M CH₃COONa solution is mixed with 25 mL of 0.02 M CH₃COOH solution the pH of the resultant solution is found to be equal to 5. The value of x is______

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answer is 10.

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Detailed Solution

Buffer soln :
$\begin{array}{l} pH = pKa + \log \frac{[salt]}{[acid]} \\ 5 = pKa + \log \frac{25 \times 0.2}{25 \times 0.02} \Rightarrow pKa = 5 - 1 = 4 \end{array}$
$\Rightarrow$ Ka = 10^-4 = 10 x 10^-5
= 10

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