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

#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"#