Answer – The pH of 1M NH₃ solution is found to be 12.12.

Explanation: 

The dissociation reaction of an aqueous solution of NH₃ is as follows:

${\mathrm{NH}₃}_{\left(\mathrm{aq}\right)} + {\mathrm{H}₂\mathrm{O}}_{\left(\mathrm{l}\right)} \rightleftharpoons {\mathrm{NH}₄⁺}_{\left(\mathrm{aq}\right)} + {\mathrm{OH}⁻}_{\left(\mathrm{aq}\right)}$

Ammonia when dissolved in water forms a weak basic solution, causing NH₃ to dissociate poorly into ammonium ions (NH₄⁺) and hydroxyl ions (OH⁻). Due to the weak dissociation, the equilibrium concentration of NH₃ ions ends up greater than that of (NH₄⁺) and (OH⁻).

Let’s assume that of the 1M of aqueous NH₃, x M of aqueous NH₃ dissociates to form x M of NH₄⁺ and x M of OH⁻.

Then, the concentrations of the reactants and products can be noted in an ICE table as shown below:

For the dissociation of a weak base given by the reaction ${\mathrm{B} }_{\left(\mathrm{aq}\right)}+ {\mathrm{H}₂\mathrm{O}}_{\left(\mathrm{l}\right)} \rightleftharpoons {\mathrm{BH}⁺}_{\left(\mathrm{aq}\right)} + {\mathrm{OH}⁻}_{\left(\mathrm{aq}\right)}$, the formula for ${\mathrm{K}}_{\mathrm{b}} = \frac{\left[{{\mathrm{BH}}^{+}}_{\left(\mathrm{aq}\right)}\right]\left[{{\mathrm{OH}}^{–}}_{\left(\mathrm{aq}\right)}\right]}{\left[{\mathrm{B}}_{\left(\mathrm{aq}\right)}\right]}$.

This formula can also be applied to the dissociation of aqueous NH₃:

${\mathrm{K}}_{\mathrm{b}} = \frac{\left[{{{\mathrm{NH}}_4}^{+}}_{\left(\mathrm{aq}\right)}\right]\left[{{\mathrm{OH}}^{–}}_{\left(\mathrm{aq}\right)}\right]}{\left[{{\mathrm{NH}}_3}_{\left(\mathrm{aq}\right)}\right]}$

The given ${\mathrm{K}}_{\mathrm{b}}$ of NH₃ and the equilibrium concentration values from the ICE table can now be substituted in the above equation:

$1.8 \times {10}^{–5} = \frac{\left[\mathrm{x}\right]\left[\mathrm{x}\right]}{[1 – \mathrm{x}]}$

Since the concentration of the dissociated NH₃ is very less in comparison with the original concentration, $1 – \mathrm{x} \approx 1$.

$\Rightarrow 1.8 \times {10}^{–5} = \frac{{\mathrm{x}}^2}{\left[1\right]}$

$\Rightarrow 1.8 \times {10}^{–5} = {\mathrm{x}}^2$

$\Rightarrow \mathrm{x} = \sqrt{1.8 \times {10}^{–5}} = 0.0134 \mathrm{M}$

As x M represents the concentration of OH⁻ and NH₄⁺ in the solution:

$\Rightarrow \left[{{\mathrm{OH}}^{–}}_{\mathrm{aq}}\right] = \left[{{\mathrm{NH}}_4}^{+}\right] = 0.0134 \mathrm{M}$

$\left[{{\mathrm{OH}}^{–}}_{\mathrm{aq}}\right]$ can now be used to find the pOH of NH₃:

$\mathrm{pOH} = – \log \left[{{\mathrm{OH}}^{–}}_{\mathrm{aq}}\right]$

$\Rightarrow \mathrm{pOH} = – \log [0.0134]$

$\Rightarrow \mathrm{pOH} = 1.8723$

The pOH value can now be used to find the pH of the given NH₃ solution. Since NH₃ is a base, it’s pH ought to be greater than 7.

$\mathrm{pH} + \mathrm{pOH} = 14$

$\Rightarrow \mathrm{pH} = 14 – \mathrm{pOH}$

$\Rightarrow \mathrm{pH} = 14 – 1.8723$

$\Rightarrow \mathrm{pH} = 12.12$

Thus, the pH of 1M NH₃ = 12.12.

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