Question

The lactate ion, C3H5O3- is a weak base. A 0.40M solution has a pH of 8.728. a. Calculate the kb of the lactate ion? b. Calculate the Ka for lactic acid, HC3H5O3?

Answer 1

Consider the equilibrium,

`C_3H_5O_3^(-) rightleftharpoons HC_3H_5O_3 + OH^(-)`

where,

`K_"b" = ([HC_3H_5O_3][OH^-])/([C_3H_5O_3^-])`

If,

`"pH"_"eq" = 8.728`

Then,

`[OH]_"eq" = [HC_3H_5O_3]_"eq" = 10^(-14+8.728) approx 5.35*10^-5"M"`

Now, consider the ICE table which conforms with the preceding data,

`C_3H_5O_3^(-) rightleftharpoons HC_3H_5O_3 + OH^(-)`

Hence,

`K_"b" = ([HC_3H_5O_3][OH^-])/([C_3H_5O_3^-]) approx 7.1*10^-11`

`———————————————`

Now, consider the equilibrium,

`HC_3H_5O_3 rightleftharpoons H^(+) + C_3H_5O_3^(-)`

where,

`K_"a" = ([H^+][C_3H_5O_3^-])/([HC_3H_5O_3])`

Fortunately, recall,

`K_"w" = K_"a"K_"b" = 1.0*10^-14`

Hence,

`K_"a" = K_"w"/K_"b" approx 1.4*10^-4`