A sample of 01687 g of an unknown monoprotic acid was


A sample of 0.1687 g of an unknown monoprotic acid was dissolved in 25.0mL of water and titrated with 0.1150 M NaOH. The acid required 15.5mL of base to reach the equivalence point. (a) Explain what is the molecular weight of the acid? (b) After 7.25mL of base had been added in the titration, the pH was found to be 2.85. Explain what is the Ka for the unknown acid?

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Chemistry: A sample of 01687 g of an unknown monoprotic acid was
Reference No:- TGS0928070

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