Question

a) A solution containing 0.0267 M maleic acid and 0.038 M disodium maleate. The Ka values for maleic acid are 1.20 × 10-2 (Ka1) and 5.37 × 10-7 (Ka2).

b) A solution containing 0.0345 M succinic acid and 0.021 M potassium hydrogen succinate. The Ka values for succinic acid are 6.21 × 10-5 (Ka1) and 2.31 × 10-6 (Ka2).

Answer 1

Qa)The solution is a buffer comprising of a weak acid and its conjugate base , maleic acid and disodium maleate.

Thus the pH of that solution is calculated using Hendersen equation.and Ka1 only need to be considered for calulation.

pH = pKa + log [conjugate base]/ [acid]

= (2-log 1.2) + log (0.038)/(0.0267)

= 2.0739

b) This solution also makes a buffer, of a weak acid and its conjugate base.

Thus its pH is calculated as

pH = pKa + log [conjugate base] / [acid] and again ka1 is to be considered.

pH = (5 - log 6.21) + log (0.021)/ )(0.0345)

= 3.9912