First slide

Acidic buffer

Question

What % of the carbon in the $H_{2} {CO}_{3} - {HCO}_{3}^{-}$ buffer should be in the form of ${HCO}_{3}^{-}$ so as to have a neutral solution ? $(K_{a} = 4 \times 10^{- 7})$

Difficult

A

20%
B

40%
C

60%
D

80%

Solution

$pH = {pK}_{a} + \log \frac{[{HCO}_{3}^{-}]}{[H_{2} {CO}_{3}]} \Rightarrow 7 = 7 - \log 4 + \log \frac{[{HCO}_{3}^{-}]}{[H_{2} {CO}_{3}]} \Rightarrow \frac{[{HCO}_{3}^{-}]}{[H_{2} {CO}_{3}]} = 4 % of carbon in the form of {HCO}_{3}^{-} = \frac{[{HCO}_{3}^{-}]}{[{HCO}_{3}^{-}] [H_{2} {CO}_{3}]} \times 100 = \frac{4}{1 + 4} \times 100 = 80 %$

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