# How to determine the purity of lactic acid with titration?

$$\pu{25 mL}$$ of $$\pu{0.5 \%(w/v)}$$ lactic acid ($$\ce{C3H6O3}$$, molecular mass $$\pu{90.1 g mol-1}$$, $$\mathrm{p}K_{\mathrm{a}} = 3.86$$ was neutralised by $$\pu{13.1 mL}$$ of $$\pu{0.1010 M}$$ monobasic base. Estimate the end point $$\mathrm{pH}$$ and calculate the amount of $$\ce{C3H6O3}$$ in % of the initial one.

This question was part of my final paper. I can't remember the exact words used in the paper for the % part, but I think it is asking for the % purity. I am not sure how to do it.

I calculated the $$\mathrm{pH}$$ using the formula $$\mathrm{pH} = 0.5 (\mathrm{p}K_\mathrm{a} - \log M)$$ but I think it is wrong. How is this doen correctly?

• 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} &= \frac{1}{2}(\mathrm{p}K_\mathrm{w} + \mathrm{p}K_\mathrm{a} - \mathrm{p}C')\\ &= \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 $$P\%$$: $$P\% = \frac{m\times 100\%}{m'}=95.36\%$$