Answer to Question #290722 in Organic Chemistry for Mikee

Question

A 0.3606-g of a weak monoprotic was dissolved in water and then titrated with 0.1040 N KOH solution. A total of 28.39 mL of the base was consumed in the titration to reach the endpoint. What is the equivalent weight of the acid? If the pH at ½ EP is 4.194, what is the Ka of the acid?

Answer 1

0.3606/28.39= 0.0127

0.0127/0.104

= 0.122

1-0.122= 0.878

= 4.194+log(0.878/0.122)

= 4.194+0.86

= 5.054