Question

Vitamin C is ascorbic acid (${\mathbf{\text{HC}}}_{\mathbf{6}}{\mathit{H}}_{\mathbf{7}}{\mathit{O}}_{\mathbf{6}}$), for which K_a is$\mathbf{8}.\mathbf{0}\times {\mathbf{10}}^{-\mathbf{5}}$. Calculate the pH of a solution made by dissolving a 500-mg tablet of pure vitamin C in water and diluting it to 100 mL.

Short answer

The pH of the solution is 2.83.

Step-by-step solution

Ascorbic acid.

Vitamin C is also known as ascorbic acid. This is an essential component in the metabolism. This ascorbic acid, or vitamin C, is water soluble. All citrus fruits like lemon, and orange contains vitamin C.

Conversion of mg into moles.

Ascorbic acid tablet is 500 mg. Convert this mg into moles.

m = 0.500 mg$H{C}_6{H}_7{O}_6$Ascorbic acid = $2.84\times {10}^{-3}$

Reaction of water and ascorbic acid.

The reaction goes as;

$H{C}_6{H}_7{O}_6\left(aq\right)+H_{2}O\left(l\right)\leftrightharpoons H_3{O}^{+}\left(aq\right)+C_6{H}_7{O}_6^{-}\left(aq\right)$

The association constant;

$\begin{array}{c}{K}_a=\frac{\left(H^{+}\right)\left(A^{-}\right)}{HA}\\ K_a=\frac{\left(H_3{O}^{+}\right)\left(C_6{H}_7{O}_6^{-}\right)}{\left(H{C}_6{H}_7{O}_6\right)}\\ K_a=\frac{X^2}{2.84\times {10}^{-2}}-x\\ =8.0\times {10}^{-5}\end{array}$

Therefore, the concentration;

X= $\left[H3O^{+}\right]$

Hence;

$\left[H3O^{+}\right]=1.47\times {10}^{-3}$

And;

$\begin{array}{l}pH=-log\left[H3O^{+}\right]\\ =2.83\end{array}$