Ascorbic acid (vitamin C, C6H8O6) is a diprotic acid (Ka1 = 7.9 × 10-5, Ka2 = 1.6 × 10-12). Calculate the pH of a solution that contains 2.1 mg acid per mL water. (Assume that only the first ionization is important in determining pH.)

The pH of the solution is 3.03

Explanation:

We are given:

Concentration of acid = 2.1 mg/mL

Converting this into mol/L, we use the conversion factor:

1 g = 1000 mg

1 L = 1000 mL

Molar mass of ascorbic acid = 176.12 g/mol

Converting the concentration, we get:

The chemical equation for the first dissociation of oxalic acid follows:

Initial:            0.0119

At eqllm:       0.0119-x               x                 x

The expression of first equilibrium constant equation follows:

We know that:

Putting values in above equation, we get:

Neglecting the negative value of 'x', because concentration cannot be negative.

To calculate the pH of the solution, we use the equation:

We are given:

Putting values in above equation, we get:

Hence, the pH of the solution is 3.03

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