A 0.551 L solution of 1.37 M sulfurous acid (Ka1 = 1.5e-2 and Ka2 = 1.0e-7) is t

Question

A 0.551 L solution of 1.37 M sulfurous acid (Ka1 = 1.5e-2 and Ka2 = 1.0e-7) is titrated with 1.65 M NaOH. What will the pH of the solution be when 0.7549 L of the NaOH has been added? The answer is pH = 8.21. Could you please show work to get to this answer? A 0.551 L solution of 1.37 M sulfurous acid (Ka1 = 1.5e-2 and Ka2 = 1.0e-7) is titrated with 1.65 M NaOH. What will the pH of the solution be when 0.7549 L of the NaOH has been added? The answer is pH = 8.21. Could you please show work to get to this answer?

Explanation / Answer

First calculate moles of each acid and base = molarity (M) x volume (L)

moles of H2SO3 = M x L = 1.37 x 0.551 = 0.755 moles

moles of NaOH = 1.65 M x 0.7549 L = 1.246 moles

Thus, we will have 0.755 moles of HSO3^-1 formed and 0.378 mols of SO3^2- formed in solution

HSO3^-2 + H2O ------> H3O+ + SO3^2-

Ka2 = [H3O+][SO3^2-]/[HSO3^-1]

1 x 10^-7 = [H3O+](0.378)/(0.755)

[H3O+] = 2.0 x 10^-7 M

pH = -log[H+] = -log(2.0 x 10^-7) = 6.70

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