Question

20 mL of .10 M HC2H3O2l solution in a small beaker is titrated with .10 M NaOH solution (at 25 C). Calculate the value of the equilibrium constant for this titration. Is it safe to assume that this titration reaction goes essentially to completion? Explain.

Answer 1

note: the given acid formula is not clear enough. but

assuming that the given acid is acetic acid (CH₃COOH) the reaction is supposed to be a weak acid versus strong base which can never reach completion but attains equilibrium.

the reaction is as follows:

CH₃COOH + NaOH <======>   CH₃COONa + H₂O

according to the given data:

no.of moles of CH₃COOH = molarity x vol of solutution in ml   = 0.1 x 20 = 0.002 moles.

                                                     1000                                   1000

and no. of moles of NaOH = 0.1 x 20 = 0.002moles

                                           1000

therefore Kc = [CH₃COONa] [ H₂O]

                       [CH₃COOH] [NaOH]

                 [H₂O] = 1 Excess Quantity

hence Kc = 0.002/ 0.002x0.002 = 1/0.002 = 500