Question

A solution of HCOOH has 0.16M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium? Express the pH numerically.

Answer 1

`pH=2.28`

Step-by-step explanation:

Hello,

In this case, for the acid dissociation of formic acid (HCOOH) we have:

`HCOOH(aq)\rightarrow H^+(aq)+HCOO^-(aq)`

Whose equilibrium expression is:

`Ka=([H^+][HCOO^-])/([HCOOH])`

That in terms of the reaction extent is:

`1.8x10^(-4)=(x*x)/(0.16-x)`

Thus, solving for
`x` which is also equal to the concentration of hydrogen ions we obtain:

`x=0.00528M`

`[H^+]=0.00528M`

Then, as the pH is computed as:

`pH=-log([H^+])`

The pH turns out:

`pH=-log(0.00528M)\\\\pH=2.28`

Regards.