i have no idea what to do for this chemistry problem?

Question

The ionization constant of a certain acid, HA, is 9.0 x 10(to the negative twelth power) If enough HA is added to water to make a 1.0 M solution, what will be the hydrogen ion concentration?




An acetic acid solution contains 3 g of acetic acid, HC2H302, in .50 dm3. If Ka for acetic acid is 1.8x 10 (to the negative 5), what is the hydrogen in concentration




they are both asking the same thing i jus dont get it i know what Ka is but....please help

Answer 1

For the equilibrium
HA <> H+ + A-
initial concentration
1.0

at equilibrium
1.0-x.......x.......x

9.0 x 10 ^-12 = (x)(x) / 1-x

x= 3 x 10^-6 M = [H+]




MM(acetic acid)= 60 g/mol

3/60= 0.05 moles

Initial concentration = 0.05 /0.5 = 0.1 M

CH3COOH <> CH3COO- + H+
initial concentration
0.1 . . . . . . . . . . . .0 . . . . . .0

at equilibrium
0.1-x . . . . . . . . . . . . x. . . . . . .x

1.8 x 10^-5 = (x)(x) / 0.1-x

x= 0.00134 M = [H+]

Answer 2

HA <---> H+ + A-

Ka = [H+][A-]/[HA]

If [HA] = 1 M, then a certain amount x will dissociate to form H+ and A- such that the Ka expression is met. Since 1 M >>> 9.0 * 10^-12, the amount x will be neglible compared to 1 M.

9.0 * 10^-12 = x^2 / 1.0
x = 3.0 * 10^-6 = [H+] pH = 5.52

3 g HC2H3O2 / 60 g/mole = 0.05 moles
0.05 moles / 0.500 L = 0.10 M

1.8 * 10^-5 = x^2/0.10 -x

Marginal if x<<<0.10, but we'll make the assumption

x^2 = 1.8 * 10^-6
x = 0.00134 M = [H+] pH = 2.87