Question

A 0.100 mole quantity of a monoprotic acid HA is added to 1.00 L of pure water. When equilibrium is reached, the pH of the solution is 3.75. What is the value of Ka for the acid HA

Answer 1

`3.167* 10^(-7)` is the value of
`K_a`for the acid HA.

Step-by-step explanation:

The pH of the solution = 3.75

The hydrogen ion concentration :

`pH=-log[H^+]`

`3.75=-\log[H^+]`

`[H^+]=10^(-3.75)=0.0001778 M`

Concentration of weak acid = 0.100 mole/L

`HA\rightleftharpoons H^++A^-`

Initually

0.100 M 0 0

At equilibrium

(0.100-x)M x x

The expression of dissociation constant can be given as:

`K_a=(x* x)/((0.100-x))`

Here the value of x =
`[H^+]=0.0001778 M`

`K_a=(0.0001778 M* 0.0001778 M)/((0.100-0.0001778)M )`

`K_a=3.167* 10^(-7)`

`3.167* 10^(-7)` is the value of
`K_a`for the acid HA.