Calculating pH for titration of weak base with strong acid
First of all, when you titrate  a weak base (methylamine) with a strong acid, the equation of titration is:$$\ce{H+(aq) + CH3NH2 -&gt; CH3NH3+ }$$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:
$$\ce{CH3NH3+&lt;=&gt;H+(aq) + CH3NH2}$$ 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: $\mathrm{pH}= 5.94$