Question

The$\mathrm{pH}$ value of the decinormal solution of ${\mathrm{NH}}_4\mathrm{OH}$, which is $20\%$ ionised, is:

Answer 1

${\mathrm{NH}}_4\mathrm{OH}$:

• ${\mathrm{NH}}_4\mathrm{OH}$ is ammonium hydroxide.
• It is a weak base.

Decinormal solution:

• The decinormal solution is a term used to describe a solution in which one litre of the substance's solution contains one-tenth $(1/10\mathrm{th})$ gram equivalent mass of the substance. It is designated as an$\mathrm{N}/10\mathrm{or}0.1\mathrm{N}$solution.
• It means the concentration is$0.1\mathrm{N}.$

Step 1: Dissociation of ${\mathrm{NH}}_4\mathrm{OH}$

• The ammonium hydroxide will be associated as:

${\mathrm{NH}}_4\mathrm{OH}\left(\mathrm{aq}\right)\rightleftharpoons {{\mathrm{NH}}_4}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

• It is a weak base, it dissociates partially.

Step 2: Finding the concentration of ${\mathrm{OH}}^{-}$

• The concentration of ${\mathrm{OH}}^{-}$is:

$$\begin{gathered} \mathrm{\alpha }(\mathrm{ionisation}\mathrm{of}{\mathrm{OH}}^{-})=\frac{20}{100} \\ =0.2 \\ \left[{\mathrm{OH}}^{-}\right]=\mathrm{C\alpha }(\mathrm{where}\mathrm{C}\mathrm{is}\mathrm{concentratonn}\mathrm{of}{\mathrm{OH}}^{-}\&\mathrm{\alpha }\mathrm{is}\mathrm{degree}\mathrm{of}\mathrm{ionisation}) \\ =0.1\times 0.2\mathrm{M} \\ =0.02\mathrm{M} \end{gathered}$$

Step 3: Finding $\mathrm{pH}$
$$\begin{gathered} \mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right] \\ =-\log [0.02] \\ \approx 1.7 \\ \mathrm{pH}=14-\mathrm{pOH} \\ =14-1.7 \\ \mathrm{pH}=12.3 \end{gathered}$$