5
$\begingroup$

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 $\pu{0.0650 M}$.

$\endgroup$
3
$\begingroup$

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: $\mathrm{pH}= 5.94$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.