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

Answer:

$pH=2.28$

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:

$Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}$

That in terms of the reaction extent is:

$1.8x10^{-4}=\frac{x*x}{0.16-x}$

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.