Question

calculate the benzoic acid, C6H5COOH, needed to provide a [H3O+] of 3.5 x 10^-4 M. the Ka of benzoic acid is 6.3 x 10^-5.

Answer 1

For the reaction:

C6H5COOH + H2O C6H5COO- + H3O+

ICE table :

	C6H5COOH	C6H5COO-	H3O+
Initial concnetration, M	C	0	0
change	-x	+x	+x
equillibrium concnetration, M	C-x	x	x

at equillibrium Ka = [C6H5COO- ][H3O+]/[C6H5COOH]

given that Ka = 6.3*10-5

using ICE table we get:

6.3*10-5 = x*x/C-x

as [H3O+] = 3.5*10-4 M = x

thus: 6.3*10-5 = (3.5*10-4)2/(C-3.5*10-4)

C-3.5*10-4 = (3.5*10-4)2/(6.3*10-5)

C-3.5*10-4 = 1.94*10-3

C = 2.29*10-3 M

2.3*10-3 M benzoic acid, C6H5COOH is needed to provide a [H3O+] of 3.5 x 10-4 M.