Calculate the pH of 0.0070 M butanoic acid, which is a monoprotic acid; Ka=1.52 x 10−5. Assume that the 5% approximation rule applies. Provide your answer to two places after the decimal?

Question

10814 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-09T03:22:01

$color(blue)"pH=3.50"$

Butanoic acid ( $HC_4H_7O_2$ ) is a weak acid .

DISSOCIATION

$HC_4H_7O_2 <=>. H^+ + C_4H_7O_2 ^-$

The equilibrium constant for this expression is called the acid dissociation constant, Ka.

$Ka$ = $1.52" x"10^-5$

The expression for Ka is:

$Ka$ = $([C_4H_7O_2 ^-] [H^+]) / ([HC_4H_7O_2])$

AT EQUILIBRIUM
Write each equilibrium concentration in terms of x

$x$ = $[C_4H_7O_2 ^-]$ = $[H^+]$

$C$ = $[HC_4H_7O_2]$

So,

$K_a$ = $([x] [x]) / ([C - x])$ = $(x^2) / (C - x)$

With the small x approximation, $C - x ~~ C$ so that:

$x ~~ sqrt(CK_a) = 3.187xx10^-4$ mol/L

Calculating,

$pH=-log[H^+]$

$pH=-log[3.187" x"10^-4]$

$color(blue)"pH=3.50"$

NOTES

Checking the 5% Rule (not used in here)

$("3.187 E-4"/0.007)100 = 4.6% < 5%$

$=>$ $"5% Rule should be considered"$

$"x-solution negative value was not considered"$