Calculate pH given mL and molarity!! please help!!

how do i calculate the pH of a weak acid? I was doing ICE for the equations but I'm getting the same answer for 1&2 and then 3&4...

1) 1.0x10^-1: 5mL of 0.10M HA
2)1.0x10^-2: 1mL of 0.10 M HA +9mL H20
3)1.0x10^-3: 5mL 0.0010M HA
4)1.0x10^-4: 1mL of 0.0010M HA +9mL H20

Ka= 1.8x10^-5


thank you!!!

Construct an ICE chart:

Molarity . . . . . .HA + H2O <==> H3O+ + A-
initial . . . . . . . .0.10 . . . . . .. . . . . . .0 . . . . . 0
change . . . .. . . .-x . . . . . . ... . . . . .x . . . . . .x
at equilibrium . .0.10-x . . . . . . . .. . . .x . . . . . x

Ka = [H3O+][A-] / [HA] = x^2 / (0.1-x) = 1.8 x 10^-5
Because x will be small compared to 0.1, we can neglect the -x term.
x^2 / 0.1 = 1.8 x 10^-5
x^2 = 1.8 x 10^-6
x = 1.3 x 10^-3 = [H3O+]

pH = -log [H3O+] = -log (1.3 x 10^-3) = 2.87

Here's what it would look like for 0.01 M HA. Again we can neglect the -x term. You can neglect the -x term if [HA] / Ka is greater than 100.

Ka = x^2 / (0.01-x) = 1.8 x 10^-5
x^2 / 0.01 = 1.8 x 10^-5
x^2 = 1.8 x 10^-7
x = 0.00042 = [H3O+]

pH = -log (0.00042) = 3.37

In the other cases, the value of -x will NOT be negligible to the molarity of HA and we will have to solve a quadratic equation.

For [HA] = 0.0010 M

Ka = x^2 / (0.0010-x) = 1.8 x 10^-5
x^2 = (1.8 x 10^-5)(0.0010 - x)
x^2 = -1.8 x 10^-5x + 1.8 x 10^-8
x^2 + 1.8 x 10^-5x - 1.8 x 10^-8 = 0 . . .using the quadratic formula, we get a positive root of

x = 0.00013 = [H3O+]
pH = -log (0.00013) = 3.89

For [HA] = 0.00010 M, x = 0.000034 = [H3O+] and pH = 4.47.