# Neutralisation and end point pH

So I just finished my final paper yesterday and I tried to attempt this 8 mark question. I am not sure how to do it.

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

(I can’t remember the exact words used in the paper for the % part, but I think it is asking for like % purity?)

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. Any help would be appreciated.

• At the end point, you have a solution of sodium lactate (weak base). The concentration of sodium lactate is given by the equation: $$C'=\frac{C_b\, V_b{eq}}{V_a +V_b{eq}}$$ Where $C_b$ is the concentration of the base,$V_b{eq}$ is the voulume of base at the end point and $V_a$is the voulume of lactic acid. $$C'=\frac{0.1010 \times 13.1}{25 + 13.1}= 0.03473 \,\ce{mol/L}$$ The $\ce{pH}$ of the solution at the end point is given by the equation: $$\ce{pH}= \frac{1}{2}(\ce{pK_w +pK_a -pC'})$$ $$\ce{pH}= \frac{1}{2}(14 +3.86 +\log 0.03473 )=8.20$$
• To calculate the purity percentage of lactic acid, we calculate the number of moles of pure lactic acid: At the end point $$n_a=n_b$$ This means: $$n_a=C_b\,V_b= 0.1010\times 13.1 \times 10^{-3 }=0.001323 \mathrm {moles}$$ The mass of pure lactic acid is: $$m=0.001323 \times 90.1= 0.1192\, \mathrm{g}$$ Or the mass of $0.5 \% w/v$ lactic acid is: $$m'=\frac {0.5 \times 25}{100}= 0.125 \,\mathrm{g}$$ The purity percentage of lactic acid $P \%$: $$P\%=\frac{m\times 100}{m'}=95.36\%$$