Question

What is the ratio of [a–]/[ha] at ph 5.75? the pka of formic acid (methanoic acid, h–cooh) is 3.75?

Answer 1

This problem can be solved using henderson equation, that is:

$pH= pKa + log\frac{[A^{-1}]}{[HA]}$

$pH - pKa = log\frac{[A^{-1}]}{[HA]}$

$5.75-3.75= log\frac{[A^{-1}]}{[HA]}$

$2=log\frac{[A^{-1}]}{[HA]}$

$\frac{[A^{-1}]}{[HA]}=10^{2}$

So, the ratio of $\frac{[A^{-1}]}{[HA]}$ is 10².

Answer 2

Answer : The ratio of $\frac{[a–]}{[ha]}$ is, 100

Solution : Given,

pH = 5.75

$pK_a$ = 3.75

The equilibrium reaction of methanoic acid is,

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

Formula used :

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

or,

$pH=pK_a+\log \frac{[a^-]}{[ha]}$

Now put all the given values in this formula, we get

$5.75=3.75+\log \frac{[a^-]}{[ha]}$

$\frac{[a^-]}{[ha]}=100$

Therefore, the ratio of $\frac{[a–]}{[ha]}$ is, 100