How would diluting HA from 0.15 M to 0.005 M affect the pH of the solution? #K_a = 1.0 xx 10^-8#.

1 Answer
Al E.
May 6, 2018

Consider the acid dissociation equilibrium,

#HA rightleftharpoons H^(+) + A^(-)#, where

#K_"a" = ([H^+][A^-])/([HA]) approx 1.0*10^-8#

Recall,

#"p"H = "p"K_"a" + log(([A^-])/([HA]))#

Without doing any arithmetic, I could tell you the relationship between the concentration of the parameters you're asking about from the preceding equation.

#"p"H propto log(1/([HA]))#

To be sure, #"p"H# is proportional to the negative logarithm of #[HA]#.

Hence, #"p"H# would generally rise in response to diluting the solution of the weak acid.

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