Question

Determine the pH of a solution that is 0.15 M HClO2 (Ka = 1.1 x 10-2) and 0.15 M HClO (Ka = 2.9 × 10-8).

Answer 1

pH = 1.39

Step-by-step explanation:

Given that:

The molarity of
`HClO_2` = 0.15 M

with acid dissociation constant
`K_a` =
`1.1 * 10^(-2)`

Acid dissociation constant of HClO =
`2.9 * 10^(-8)`

Molarity of HClO = 0.15 M

The objective is to determine the pH of the solution:\

To determine the concentration of
`H_3O^+` obtained from both acids

The equation for the reaction can be expressed as :

`HClO_2 + H_2O \to H_3O^+ + ClO_2^-`

The dissociation constant for the above reaction is as follows:

`ka = \frac{ [H_3O^+] [ClO_2^-]} { [HClO_2]}`

`1.1 * 10^(-2) = \frac{ [x] [x]} { [0.15]}`

`1.1 * 10^(-2) = \frac{ [x]^2 } { [0.15]}`

`x^2 = 0.15 * 1.1 * 10^(-2)`

`x^2 = 0.00165`

`x=√( 0.00165)`

x = 0.04062 M

Now to determine the concentration of
`H3O^+` obtained from HClO

The equation for the reaction can be expressed as :

`HClO + H_2O \to H_3O^+ + ClO^-`

`ka = ([H3O^+] [ClO^-])/( [HClO])`

`2.9 * 10 ^(-8) = ([x] [x])/( [0.15])`

`2.9 * 10 ^(-8) * [0.15] = {[x]^2}`

`{[x]} ^2=4.35 * 10^(-9)`

`x=\sqrt{4.35 * 10^(-9)}`

`x=6.595 * 10^(-5)`

Thus the total concentration now is :

x =
`H_3O^+` =
`0.04062 + 6.595 * 10^(-5) M`

`H_3O^+` = 0.04068595 M

`H_3O^+`
`\simeq` 0.04069 M

pH = -log [H3O⁺]

pH = -log [0.04069]

pH = 1.39