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?

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Answer 1 · 2018-04-09T03:22:01

$pH=3.50$ $color(blue)"pH=3.50"$

Butanoic acid ( $H C_{4} H_{7} O_{2}$ $HC_4H_7O_2$ ) is a weak acid .

DISSOCIATION

$H C_{4} H_{7} O_{2} \Leftrightarrow . H^{+} + C_{4} H_{7} O_{2}^{-}$ $HC_4H_7O_2 <=>. H^+ + C_4H_7O_2 ^-$

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

$K a$ $Ka$ = $1.52 x 10^{- 5}$ $1.52" x"10^-5$

The expression for Ka is:

$K a$ $Ka$ = $\frac{[C_{4} H_{7} O_{2}^{-}] [H^{+}]}{[H C_{4} H_{7} O_{2}]}$ $([C_4H_7O_2 ^-] [H^+]) / ([HC_4H_7O_2])$

AT EQUILIBRIUM
Write each equilibrium concentration in terms of x

$x$ $x$ = $[C_{4} H_{7} O_{2}^{-}]$ $[C_4H_7O_2 ^-]$ = $[H^{+}]$ $[H^+]$

$C$ $C$ = $[H C_{4} H_{7} O_{2}]$ $[HC_4H_7O_2]$

So,

$K_{a}$ $K_a$ = $\frac{[x] [x]}{[C - x]}$ $([x] [x]) / ([C - x])$ = $\frac{x^{2}}{C - x}$ $(x^2) / (C - x)$

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

$x \approx \sqrt{C K_{a}} = 3.187 \times 10^{- 4}$ $x ~~ sqrt(CK_a) = 3.187xx10^-4$ mol/L

Calculating,

$p H = - log [H^{+}]$ $pH=-log[H^+]$

$p H = - log [3.187 x 10^{- 4}]$ $pH=-log[3.187" x"10^-4]$

$pH=3.50$ $color(blue)"pH=3.50"$

NOTES

Checking the 5% Rule (not used in here)

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

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

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