Calculate the pH at the equivalence point for the titration of $\pu{0.130 M}$ methylamine ($\ce{CH3NH2}$) with $\pu{0.130 M}$ $\ce{HCl}$. The $K_\mathrm{b}$ of methylamine is ${5.0 \cdot 10^{–4}}$.

So I started with the equation:

$$\ce{HCl + CH3NH2 <=> CH3NH3+ + Cl-}$$

and then I knew that

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{\ce{[base]}}{\ce{[acid]}} \right)$$

So, I put ${\log \left(\frac{0.130}{0.130}\right) = \log 1 = 0}$ and then added that to the $\mathrm{p}K_\mathrm{a}$, which I got from the equation

$$\mathrm{p}K_\mathrm{a} = \frac{K_\mathrm{w}}{K_\mathrm{b}} \quad \rightarrow \quad \mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a})$$

However, after I plugged those in to get a $\mathrm{pH}$, it turned out to be wrong and then comments said that when titrated a weak base with a strong acid, the volume is doubled at equivalence point and the concentrations are halved.

Why is this? I now know that my original equation was wrong and it should be

$$\ce{CH3NH3+ <=> H+ + CH3NH2}$$

and from there I should make an ICE table with the concentration of $\pu{0.0650 M}$.


First of all, when you titrate a weak base (methylamine) with a strong acid, the equation of titration is:$$\ce{H+(aq) + CH3NH2 -> CH3NH3+ }$$The reaction of titration, as you can see is total and quantitative. The constant of the aforementioned equilibrium is: $$ K=\frac{K_\mathrm b}{K_\mathrm w}= 5\times 10^{10} \gg 10^3$$ According to the stoichiometry of the neutralization equation, at the equivalence point: $n_{\mathrm{acid}}=n_{\mathrm{base}}$ $$ C_{\mathrm{acid}}.V_{\mathrm{acid}}=C_{\mathrm{base}}.V_{\mathrm{base}}$$

As the initial concentration of the base equals the initial concentration of the acid, the volume of the acid at the equivalence point equals the volume of the base: $$ V_{\mathrm{acid}}^{equiv.}=V_{\mathrm{base}}$$ You can see that the total volume is doubled.

Now, let's calculate the $\mathrm{pH}$ at the equivalence point: We have a weak acid at the initial concentration $$C_{\ce{CH3NH3+}}=\frac{C_{\mathrm{base}}.V_{\mathrm{base}}}{V_{\mathrm{base}}+V_{\mathrm{acid}}}=\frac{C_{\mathrm{base}}.V_{\mathrm{base}}}{2V_{\mathrm{base}}}=\frac{C_{\mathrm{base}}}{2}= \frac{0.13}{2} = \pu{0.065 M}$$ It's in equilibrium with its conjugated base: $$\ce{CH3NH3+<=>H+(aq) + CH3NH2}$$ The acid is partially dissociated:$$ [\ce{CH3NH3+}]= 0.065-x$$ $$ [\ce{CH3NH2}]=x$$ $$ [\ce{H+(aq)}]=x$$ Let's write the constant of this equilibrium: $$K_\mathrm a= 0.2\times 10^{-10}= \frac{x^2}{0.065-x}$$ We solve this equation of second order to find $x$, the concentration of ion hydronium. We find: $\mathrm{pH}= 5.94$

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