Question

The weak acid HQ has pKa of 4.89.

1)Calculate the [H3O+] of .035 M HQ.

2)Calculate the (OH-) of 0.500 M HQ
....Am I correct?

a) pKa = 4.89

=> - log(Ka) = 4.89

=> Ka = antilog(- 4.89) = 10-4.89 = 1.29x10-5

1. Given concentration of HQ, C = 0.035 M

Dissociation of HQ is given as

HQ(aq) -- > H3O+(aq) + Q-(aq), Ka = 1.29x10-5

Initial conc:0.035 M 0 0

eqm conc: (0.035-x)M x M x M

Ka =([H3O+(aq)]x[Q-(aq)]) / [HQ(aq)] = (x M * x M) / (0.035-x)M

Since x < < 1, (0.035-x) is nearly equals to 0.035

=> 1.29x10-5 = x2 / 0.035

=> x = Underroot (0.035 x 1.29x10-5 ) = 6.72x10-4

x = [H3O+(aq)] = 6.72x10-4 (answer)

B)for 0.500M HQ:
k_a=(x.x )/(0.5-x)=¡¼10¡½^(-4.89)=1.26*¡¼10¡½^(-5)
x=0.00251
pH= -log0.00251=2.6

But
pH+pOH=14

pOH=11.4
[OH^- ]=¡¼10¡½^11.4=2.5*¡¼10¡½^11

Thank you.

Answer 1

a is ok.
b I obtained 2.54 for pH with pOH 11.46