Question

It takes 4.71 mL of 0.0410 M NaOH to titrate a 50.00-mL sample of flat (no CO2) GG’s Cola to a pH of 4.9. At this point the addition of one more drop (0.02 mL) of NaOH raises the pH to 6.1. The only acid in GG’s Cola is phosphoric acid. Compute the concentration of phosphoric acid in this cola. Assume that the 4.71 mL of base removes only the first hydrogen from the${\mathbf{H}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}$; that is, assume that the reaction is

${\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{H}}_2{\mathrm{PO}}_4\left(\mathrm{aq}\right)$

Short answer

The raise of$\mathrm{pH}$indicate the endpoint of the titration. So, the volume at the first equivalence point is:

$\begin{array}{l}{\mathrm{V}}_{\mathrm{e}1}=4.71\mathrm{mL}+0.02\mathrm{mL}\\ {\mathrm{V}}_{\mathrm{e}1}=4.73\mathrm{mL}\end{array}$

Step-by-step solution

Step 2

The n of is

$\begin{array}{l}\mathrm{n}\left(\mathrm{NaOH}\right)=\left[\mathrm{NaOH}\right].{\mathrm{V}}_{\mathrm{eI}}\\ \mathrm{n}\left(\mathrm{NaOH}\right)=0.0410\mathrm{M}.4{.73.10}^{-3}\mathrm{L}\\ \mathrm{n}\left(\mathrm{NaOH}\right)=1{.94.10}^{-4}\mathrm{mol}\end{array}$

At the equivalence point, the n of the base is equal to the n of acid:

$\mathrm{n}\left(\mathrm{NaOH}\right)=\mathrm{n}\left({\mathrm{H}}_3{\mathrm{PO}}_4\right)$

Step 3

The concentration of ${\mathrm{H}}_3{\mathrm{PO}}_4$

$\begin{array}{l}{\mathrm{H}}_3{\mathrm{PO}}_4=\frac{\mathrm{n}\left({\mathrm{H}}_3{\mathrm{PO}}_4\right)}{\mathrm{V}\left(\mathrm{sample}\right)}\\ {\mathrm{H}}_3{\mathrm{PO}}_4=\frac{1{.94.10}^{-4}\mathrm{mol}}{50{.10}^{-3}\mathrm{L}}\\ {\mathrm{H}}_3{\mathrm{PO}}_4=3{.88.10}^{-3}\mathrm{M}\end{array}$

Result

The reaction is${\mathrm{H}}_3{\mathrm{PO}}_4=3{.88.10}^{-3}\mathrm{M}$