Q.

A one litre solution contains 0.08 mole of acetic acid (Ka = 1.75 × 10–5). To this solution, 0.02 mole of NaOH is added. Then the pH of resulting solution is [log 1.75 = 0.243]

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a

4.28

b

4.456

c

5.058

d

5.234

answer is C.

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Detailed Solution

Following reaction will occur

 

CH3COOH

\large +

NaOH

\large \rightarrow CH3COONa \large + H2O

t = 0 0.08   0.02          0             0
\large t=t_\infty 0.08 - 0.02   0.02 - 0.02 = 0       0.02          0
  =0.06   = 0  

PH of an acidic buffer is given as

\large {P^H} = {P^{{K_a}}} + \log \frac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}

\large {P^H} = 4.74 + \log \frac{{0.02}}{{0.06}} = 4.28

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