Question

Determine the pH of a solution that is 0.15 M N*H_{3} and 0.25 M N*H_{4}*Cl

Answer 1

The pH of the buffer solution is 13.12.

The equilibrium equation for the dissociation of
`NH_4OH` is:

`NH_4OH + H_2O` ⇌
`NH_4^+ + OH^-`

Define the initial concentrations.

The initial concentrations of
`NH_4OH` and
`NH_4Cl` are:

`[NH_4OH]` = 0.15 M

`[NH_4Cl]` = 0.25 M

Set up an ICE table.

Species Initial concentration (M) Change in concentration (M) Equilibrium concentration (M)

[
`NH_4OH`] 0.15 -x 0.15 - x

[
`NH_4^+`] 0.25 +x 0.25 + x

[
`OH^-`] 0 +x x

Substitute the equilibrium concentrations into the equilibrium equation.

Kb = [
`NH_4^+`][
`OH^-`] / [
`NH_4OH`]

1.8 ×
`10^(-5)` = (0.25 + x)(x) / (0.15 - x)

Solve for x.

This equation can be solved for x using the quadratic formula. The value of x is:

x = 7.21 ×
`10^(-3)` M

Calculate the pOH and pH.

pOH = -log10([OH−])

pOH = -log10(7.21 ×
`10^(-3)`)

pOH = 2.14

pH = 14 - pOH

pH = 14 - 2.14

pH = 11.86

Question:-

Calculate the pH of the buffer solution containing 0.15 mole of
`NH_4OH` and 0.25 mole of
`NH_4Cl`. Kb for
`NH_4OH` is 1.8 ×
`10^(-5)`