Diluting a sample of weak base before titration. What happens to equivalence point pH of diluted vs undiluted?


> 
> A student titrated a 10.00 mL of 0.20 M methylamine ($\ce{CH3NH2}$) and determined that 10.00 mL of $\ce{HCl}$ was needed to reach the equivalence point.
> 
> 
> If you diluted your original sample of methylamine by adding 10.00 mL of water to your sample before you started adding $\ce{HCl}$, which of the following statements would be true?
> 
> 
> **a**) You now only need to add 5.00 mL of $\ce{HCl}$ to reach equivalence.  
> 
> **b**) The equivalence point pH would be the same as in the undiluted titration.  
> 
> **c**) Both (**a**) and (**b**) are true.  
> 
> **d**) Neither (**a**) nor (**b**) are true.
> 
> 
> 


The answer is **d**) Neither (**a**) nor (**b**) are true.


I always thought that both the $\mathrm{p}K\_\mathrm{a}$ and equivalence $\mathrm{pH}$ of an acid stay the same no matter what the concentration. Except where it is so dilute, ionization from water would have an effect. But I don't understand why or how it would change. I can't find anything in the textbook or online, can someone please explain this?




At the point of equivalence, when the added molar amount of $\ce{HCl}$ is just matching the present molar amount of $\ce{CH3NH2}$, there is solution of the weak acid $\ce{CH3NH3+}(+\ce{Cl-})$, that I will denote as $\ce{HA}$.


If we neglect $\ce{H+}$ from water dissociation and consider $c\_\ce{A-}\ll c\_\ce{HA}$, then $c\_\ce{A-} = c\_\ce{H+}$
Therefore


$$K\_\mathrm{a} = \frac{c\_\ce{H+} c\_\ce{A-}}{c\_\ce{HA}} = \frac{c\_\ce{H+} c\_\ce{H+}}{c\_\ce{HA} + c\_\ce{A-}}$$


$$\begin{align}
\mathrm{pH} &= \frac12(\mathrm{p}K\_\mathrm{a} - \log c) \\
 &= \frac12(14 - \mathrm{p}K\_\mathrm{b} - \log c) \\
 &= 7 - \frac12(\mathrm{p}K\_\mathrm{b} + \log c) \\
\end{align}$$


where $\mathrm{p}K\_\mathrm{b}$ is the methylamine basicity constant;  

$c$ is summary concentration of methylamine and it's a conjugate acid.


Therefore, if there is final volume 30 instead of 20 ml, the final


$$\Delta\mathrm{pH} = -0.5\log \left(\frac{\pu{20 ml}}{\pu{30 ml}}\right) = +0.088$$ 


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