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In what volume of 0.02M of HCl aqueous solution should be added to 0.02M of imidazole aqueousolution (base ionization constant $= 9 .8 \times 10^{- 8}$ at 250C) to make 100ml of a buffer with (pH=7). (log 9.8=0.9912) and (log(1.019)=0.008)

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33mL

74mL

26mL

67ML

answer is B.

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Detailed Solution

$pKb = - \log Kb = - \log (9 .8 \times 10^{- 8})$

=8-log9.8=(8-0.9912)=7.0088

For basic buffer solution

$P^{OH} = pKb - \log \frac{(Wb)}{(salt)}$

$(14 - 7) = 7 .008 - \log (\frac{0 .02 \times (V_{1} - V_{2})}{V_{2} \times 0 .02})$

$\log (\frac{V_{1} - V_{2}}{V_{2}}) = 0.008 \frac{V_{1} - V_{2}}{V_{2}} = 1.019 ⟶ (1) V_{1} + V_{2} = 100 ⟶ (2) V_{2} = 33 m L V_{1} = 67 mL$

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