Question

You have 100.0 g of saccharin, a sugar substitute, and you want to prepare a pH = 5.75 solution. What volume of solution can be prepared? For saccharin (HC₇H₄NSO₃), pK_a = 11.70 (pK_a = -log K_a).

Short answer

The volume required for 100.0 g of saccharin, a sugar substitute, to prepare a pH = 5.75 solution with pK_a = 11.70 is 345 mL or 0.345L.

Step-by-step solution

Calculating Ka from given pKa

K_acan be calculated as

$$\begin{gathered} {\mathrm{pK}}_{\mathrm{a}}=-\log \left[{\mathrm{K}}_{\mathrm{a}}\right]=11.70 \\ \left[{\mathrm{K}}_{\mathrm{a}}\right]={10}^{-{\mathrm{pK}}_{\mathrm{a}}}={10}^{-11.70}=2\times {10}^{-12} \end{gathered}$$

Calculating [H+] from given pH

[H⁺]can be calculated as

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]=5.75 \\ \left[{\mathrm{H}}^{+}\right]={10}^{-\mathrm{pH}}={10}^{-5.75}=1.78\times {10}^{-6} \end{gathered}$$

Calculation of saccharin concentration from Ka

Saccharin (HC₇H₄NSO₃) dissociates as

$$\begin{gathered} {\mathrm{HC}}_7{\mathrm{H}}_4{{\mathrm{NSO}}_3}_{\left(\mathrm{aq}\right)\rightleftarrows }{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}+{\mathrm{C}}_7{\mathrm{H}}_4{\mathrm{NSO}}_{3\left(\mathrm{aq}\right)} \\ \mathrm{a}00 \\ \mathrm{a}-\mathrm{x}\mathrm{x}=(1.78\times {10}^{-6})1.{7810}^{-6} \\ \mathrm{a}-1.78\times {10}^{-6}1.78\times {10}^{-6}0.001 \\ {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{C}}_7{\mathrm{H}}_4{{\mathrm{NSO}}_3}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{HC}}_7{\mathrm{H}}_4{\mathrm{NSO}}_3\right]}=\frac{{\left(1.78\times {10}^{-6}\right)}^2}{\mathrm{a}-1.78\times {10}^{-6}}=2\times {10}^{-12} \\ \frac{{\left(1.78\times {10}^{-6}\right)}^2}{\mathrm{a}}=2\times {10}^{-12}\Rightarrow \mathrm{a}=\frac{{\left(1.78\times {10}^{-6}\right)}^2}{2\times {10}^{-12}}=1.584 \\ (\mathrm{taking}\mathrm{a}-1.78\times {10}^{-6}~\mathrm{a}) \end{gathered}$$

Calculating volume of saccharin from concentration

Concentration = 1.584 mol/L

$$\begin{gathered} \mathrm{Moles}\mathrm{of}\mathrm{Saccharin}=\frac{\mathrm{mass}}{\mathrm{molar}\mathrm{mass}}=\frac{100}{183}=0.546 \\ \mathrm{Concentration}=\frac{\mathrm{moles}}{\mathrm{volume}\left(\mathrm{L}\right)}=\frac{0.546}{\mathrm{volume}\left(\mathrm{L}\right)}=1.584 \\ \mathrm{volume}\left(\mathrm{L}\right)=\frac{0.546}{1.584}=0.345\mathrm{L}\mathrm{or}345\mathrm{mL} \end{gathered}$$