Calculate the pH at the equivalence point of a titration of 62 mL of 0.1 M $\ce{CH_3NH_2}$ with 0.20 M HCl. The $\ce{K_b}=4.4\cdot10^{-4}$.

At the equivalence point, the moles of CH3NH2 equals the moles of HCl. This is simple solution stoichiometry. Turns out, we require 62 mL or the CH3NH2 and 31 mL of the HCl for a total volume of 93 mL.

Now, construct an equilibrium equation for the weak base: $\ce{CH_3NH_2 + H_2O <=> HCH_3NH_2^+ + OH^-}\ \ \ \ \ \ $ or
$\ce{HCH_3NH_2^+ <=> H^+ + CH_3NH_2}$.

From here on out, I get really confused. You have to also make another equation that includes HCl to finally figure out equilibrium of $\ce{H^+}$, but I don't know how to get there.

  • $\begingroup$ I hope that these hydrogens before carbons are just "accident" and not some misconception - edit accordingly. $\endgroup$ – Mithoron Apr 4 at 22:07
  • $\begingroup$ @Mithoron It helps me see it as a proton attached to a base, rather than an entirely new compound. $\endgroup$ – Christopher Marley Apr 5 at 1:28

Edit: At the equivalence point, the solution contains the dissolved salt $\ce{CH3NH3Cl}$, dissociated to $\ce{CH3NH3+ + Cl-}$, as @MaxW noted.

You can calculate $pH$ of the conjugate acid $\ce{CH3NH3+}$ at its dissociation equilibrium.

$$\begin{align} K_w &= K_a . K_b = 10^{-14}\\ pK_w &= pK_a + pK_b = 14\\ \ce{ CH3NH3+ &<=> H+ + CH3NH2 } \end{align}$$

With the simplifying assumptions $$\begin{align} c_{\ce{H+}} &= c_{\ce{CH3NH2}} \\ c_{\ce{CH3NH2}} &<< c_{\ce{CH3NH3+}} \end{align}$$

is the case rather simple:

$$\begin{align} K_a &= \frac{ c_{\ce{H+}}.c_{\ce{CH3NH2}}}{ c_{\ce{CH3NH3+}}} = \frac{ c_{\ce{H+}}^2}{ c_{\ce{CH3NH3+}}} \\ pH &= 1/2.( pK_a - \log{c_{\ce{CH3NH3+}} } ) \\ pH &=7 - 1/2.( pK_b + \log{c_{\ce{CH3NH3+}} } ) \end{align}$$

  • $\begingroup$ Yes. The problem is essentially like you are dissolving $\ce{CH3NH3Cl}$ $\endgroup$ – MaxW Apr 4 at 23:02

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