Doubly Sour

Chemistry Level 3

A diprotic acid H 2 X \text{H}_2\text{X} has K 1 = 2.0 ⋅ 1 0 − 6 K_1 = 2.0\cdot 10^{-6} and K 2 = 5.0 ⋅ 1 0 − 8 K_2 = 5.0\cdot 10^{-8} .

If this acid is present in a solution of pH = 7 \text{pH } = 7 , determine the ratio [ X 2 − ] [ H 2 X ] . \dfrac{[\text{X}^{2-}]}{[\text{H}_2\text{X}]}.


The answer is 10.

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Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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1 solution

Arjen Vreugdenhil
Feb 10, 2016

pH = 7 implies that [ H + ] = 1 0 − 7 [\text{H}^+] = 10^{-7} .

By definition, K 1 = [ H + ] [ HX − ] [ H 2 X ] , K 2 = [ H + ] [ X 2 − ] [ HX − ] . K_1 = \frac{[\text{H}^+][\text{HX}^-]}{[\text{H}_2\text{X}]},\ \ \ K_2 = \frac{[\text{H}^+][\text{X}^{2-}]}{[\text{HX}^-]}. Multiply these two to find K 1 ⋅ K 2 = [ H + ] 2 [ X 2 − ] [ H 2 X ] . K_1\cdot K_2 = \frac{[\text{H}^+]^2[\text{X}^{2-}]}{[\text{H}_2\text{X}]}. Therefore the ratio is [ X 2 − ] [ H 2 X ] = K 1 ⋅ K 2 [ H + ] 2 = 2.0 ⋅ 1 0 − 6 ⋅ 5.0 ⋅ 1 0 − 8 ( 1 0 − 7 ) 2 = 1 0 − 13 1 0 − 14 = 10 . \frac{[\text{X}^{2-}]}{[\text{H}_2\text{X}]} = \frac{K_1\cdot K_2}{[\text{H}^+]^2} = \frac{2.0\cdot 10^{-6}\cdot 5.0\cdot 10^{-8}}{(10^{-7})^2} = \frac{10^{-13}}{10^{-14}} = \boxed{10}.

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