How to determine the purity of lactic acid with titration?
At the end point, you have a solution of sodium lactate (weak base). The concentration of sodium lactate is given by the equation:
$$
C' = 
\frac{C_\mathrm{b}\, V_\mathrm{b,eq}}{V_\mathrm{a} + V_\mathrm{b, eq}},
$$ 
where $C_\mathrm{b}$ is the concentration of the base, $V_\mathrm{b, eq}$ is the volume of the base at the end point and $V_\mathrm{a}$ is the volume of lactic acid.
$$
C' =
\frac{\pu{0.1010 mol L-1} \times \pu{13.1 mL}}{\pu{25 mL} + \pu{13.1 mL}}
 = \pu{0.03473 mol L-1}
$$ 
The $\ce{pH}$ of the solution at the end point is given by the equation:
\begin{align}
  \mathrm{pH} 
    &amp;= \frac{1}{2}(\mathrm{p}K_\mathrm{w} + \mathrm{p}K_\mathrm{a} 
      - \mathrm{p}C')\\
    &amp;= \frac{1}{2}(14 + 3.86 + \log 0.03473) = 8.20
\end{align}

To calculate the purity percentage of lactic acid, we calculate the amount of substance of pure lactic acid: At the end point 
$$n_\mathrm{a} = n_\mathrm{b}.$$ 
This means: 
$$
n_\mathrm{a} 
  = C_\mathrm{b} V_\mathrm{b}
  = \pu{0.1010 mol L-1}\times \pu{13.1E-3 L} 
  = \pu{0.001323 mol}
$$ 
The mass of pure lactic acid is: 
$$
m 
  = \pu{0.001323 mol} \times \pu{90.1 g mol-1}
  = \pu{0.1192 g}
$$
Or the mass of $\pu{0.5 \%(w/v)}$ lactic acid is:
$$
m'
  = \frac{\pu{0.5 \%(w/v)} \times \pu{25 mL}}{100\%} = \pu{0.125 g}
$$
The purity percentage of lactic acid $P\%$:
$$P\% = \frac{m\times 100\%}{m'}=95.36\%$$