Consider a the titration of 2.5 M sulfurous acid (H2SO3, Ka1 = 1.5e-2, Ka2 = 1.0e-7)...

Question

Question

Consider a the titration of 2.5 M sulfurous acid (H2SO3, Ka1 = 1.5e-2, Ka2 = 1.0e-7) with 2.0 M NaOH.

What is the pH at the first halfway point of the titration?

What is the pH at the second halfway point of the titration?

What is the pH at the equivalence point of the titration?

science chemistry

Answer 1

Answer #1

we have

Ka1 = 1.5e-2, Ka2 = 1.0e-7

pKa1 = - log Ka1 = - log (1.5 * 10^-2) = 1.82

pKa2 = - log Ka2 = - log (1.0 * 10^-7) = 7.0.

pH at the first halfway point of the titration = pKa1 = 1.82

pH at the second halfway point of the titration = pKa2 = 7.0

pH at the equivalence point of the titration = (pKa1 + pKa2) / 2 = (1.82 + 7.0) / 2 = 4.41

Answer 2

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