Question

What is the ratio [ A − ] / [ HA ] at pH 5.75 ? The p K a of formic acid (methanoic acid, H − COOH ) is 3.75.

Answer 1

Answer : The ratio of $\frac{[A^-]}{[HA]}$  at pH 5.75 is, 100

Explanation : Given,

$pK_a=3.75$

$pH=5.75$

The equilibrium reaction of methanoic acid is,

$HCOOH\rightleftharpoons HCOO^-+H^+$

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[HCOO^-]}{[HCOOH]}$

or,

$pH=pK_a+\log \frac{[A^-]}{[HA]}$

Now put all the given values in this expression, we get:

$5.75=3.75+\log \frac{[A^-]}{[HA]}$

$\frac{[A^-]}{[HA]}=100$

Therefore, the ratio of $\frac{[A^-]}{[HA]}$  at pH 5.75 is, 100