(Answered) A sample of 01687g of an unknown monoprotic acid was

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A sample of 0.1687g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.1150 M NaOH. The acid required 15.5 mL of base to reach the equivalence point.

(a) What is the molecular weight of the acid?

(b) After 7.25 mL of base had been added in the titration, the pH was found to be 2.85. What is the Ka for the unknown acid.

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