pKa = 4.89 so Ka = 10^-4.89 = 1.29x10^-5
Ka = 1.29x10^-5 = [H+][Q-] / [HQ]
1.29x10^-5 = x^2 / 0.046 - x
Shouldn't this be
1.29E-5 = x^2/0.5 and you don't need the quadratic. I get 2.54E-3 for x. pH = 2.59 but check me out on that.
x^2 + 1.29x10^-5x - 5.93x10^-7 = 0
x = [H+] = 7.64x10^-4M
pH = -log[H+] = 3.12
The weak acid HQ has pKa of 4.89. Calculate the (OH-) of 0.500 M HQ
This is what I have:
pKa = 4.89 so Ka = 10^-4.89 = 1.29x10^-5
Ka = 1.29x10^-5 = [H+][Q-] / [HQ]
1.29x10^-5 = x^2 / 0.046 - x
x^2 + 1.29x10^-5x - 5.93x10^-7 = 0
x = [H+] = 7.64x10^-4M
pH = -log[H+] = 3.12
1.29x10^-5 = x^2 / 0.5 - x
x^2 + 1.29x10^-5x - 6.71x10^-6 = 0
x = [H+] = 3.146x10^-3M
[OH-] = 1x10^-14 / 3.146x10^-4 = 3.179x10^-12M = [OH-] answer......????
1 answer
1 answer