Question

0.1 Mole Of CH₃NH₂ (K_b=5x10^-4) is mixed with 0.08 Mole of HCl and diluted to one Litre. What will be the H⁺ Concentration In the solution?

Answer 1

The reaction occurring is :
$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{NH}}_2+\mathrm{HCl}\to {\mathrm{CH}}_3{\mathrm{NH}}_2^{+}{\mathrm{Cl}}^{-} \\ 0.10.080 \\ 0.0200.08 \\ (0.1-0.08) \end{gathered}$$

The reaction mixture contains unreacted methylamine(CH₃NH₂) and hydrochloride salt(HCl).

Thus, it is a basic buffer solution.

The expression for the hydroxide ion concentration is[OH^-] = ${\mathrm{k}}_{\mathrm{b}}\times \left[\frac{\mathrm{Base}}{\mathrm{Conjugate}\mathrm{acid}}\right]$
Substitute values in the above expression= $5\times {10}^{-4}\times \left[\frac{0.02}{0.08}\right]=1.25\times {10}^{-4}$

But $\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={\mathrm{k}}_{\mathrm{W}}={10}^{-14}$

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]\left[1.25\times {10}^{-4}\right]={10}^{-14} \\ \left[{\mathrm{H}}^{+}\right]=\frac{{10}^{-14}}{\left[1.25\times {10}^{-4}\right]}=8\times {10}^{-11} \end{gathered}$$

0.1 Mole Of CH₃NH₂ (Kb=5x10^-4) is mixed with 0.08 Mole of HCl and diluted to one Litre. The H+ Concentration In the solution is 8x10^-11