An unknown weak acid with a concentration of 0.090 M has a pH of 1.80. What is the Ka of the weak acid?

Answer 1

1. Write generic equation
2. HA (aq) + H2O(l) <--> A- (aq) + H3O+ (aq)
3. Write ICE table

HA (aq) + H2O(l) <--> A- (aq) + H3O+ (aq)

I - 0.090 xxx | 0 0

C- _{1.58x10-2 mol/L}xxx |_{1.58x10-2 mol/L 1.58x10-2 mol/L}

E- 0.0748 xxx |_{1.58x10-2 mol/L 1.58x10-2 mol/L}

[H3O+] = 10^-PH

= 10^-1.80

=1.58x10^-2 mol/L

3.. solve for Ka using this formula

K_a = [A-][H3O+]

----------------

[HA]

=(1.58x10^-2)²

---------------

0.0748

K_a = 3.34x10^-3