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$\pu{25 mL}$ 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?

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  • 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 $P\%$: $$P\% = \frac{m\times 100\%}{m'}=95.36\%$$

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