20 ml of 0.1 M acetic acid is mixed with 50ml of potassium acetate. K_a of acetic acid = 1.8 × 10^–5 at 27⁰C. The concentration of potassium acetate if pH of the mixture is 4.8

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0.02M

0.1M

0.04M

0.4M

answer is B.

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Detailed Solution

$\large {P^H} = {P^{Ka}} + \log \frac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}}\$

$\large \Rightarrow 4.8 = 4.74 + \log \left( {\frac{{N \times 50}}{{0.1 \times 20}}} \right)\$

$\large Assuming4.74 \sim 4.8\$

$\large \log \left( {\frac{{N \times 50}}{{0.1 \times 20}}} \right) = 0$
$\large N \times 50 = 0.1 \times 20$
$\large N = \frac{2}{{50}}$
$\large N = 0.04M$

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