i think im more confused about why they gave you the ka of the product but youre able to use it in the equation like that.
Because you can figure out Kb if they give you Ka.
Ka X Kb = Kw (water dissociation constant) you NEED to memorize this number. It's 1.0 X 10^-14.
So: Ka X Kb = 1.0 X 10^-14 - rearrange to find Kb
Kb = 1.0 X 10^-14 / Ka - substitute in Ka
Kb = 1.0 X 10^-14 / 5.0 X 10^-10
You can rewrite this to be 10 X 10^-15 / 5.0 X 10^-10
Solve: 2 X 10^-5
Also there's a typo..should be 0.04M not 0.4M in the solution.
So:
(x)(x) / (0.04-x) = 2x10^-5
you can approximate since CN- is a weak base it won't react completely. So 0.04-x = 0.04 almost
(x^2) / (0.04) = 2x10^-5 - Multiply by 0.04
(2.0 x 10^-5) x (4.0 x 10-2) = 8.0 x 10^-7 = x^2 - Rewrite - 80 x 10^-8 - take sqr root
Just under 9 (9^2 = 81) so 8.9ish x 10^-4 = x
So in the equation [OH-] = x = 8.9ish x 10^-4
pOH = -log (8.9 x 10^-4)
pOH = Almost a full 1 pH less than 4...round to about 3.2
pH + pOH = 14
pH = 14 - pOH so pH = 14 - 3.2 = 11.8 ish