Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0146 M solution. The pH of the resulting solution is 6.14. Calculate the Ka for the acid.

Answer 1

The acid solution is 0.0146 M

so [acid] = 0.0146

It's monoprotic ... like HA ... so when it dissociates HA ? H? + A? ... so [H?] = [A?]

pH = -log [H?]

so -log [H?] = 2.54

[H?] = 10^(-2.54)

[H?] = 0.00288 = [A?]

set up the ICE table:

................. [HA] ...................... [H?] ........................ [A?]

Initial ..... 0.0146 ...................... 0 ............................. 0

change -0.00288 .............. +0.00288 ............... +0.00288

equilibrium 0.01172 ............ 0.00288 ................. 0.00288


Ka = [H?] [A?] / [HA]

Ka = 0.00288