Finding [OH] in a .35M NaCN?

Question

The Ka for hydrocyanic acid (HCN) is 4.9 X 10-10. What is [OH-] in a 0.35 M solution of sodium cyanide (NaCN)?

heres how i approached

x^2/.35=4.9*10^-10

x=1.31*10^-5

pH= -log(answer)
pH=4.88

pOH=14-4.88
pOH=9.12

[OH]= 10^-9.12

im doing something wrong.

Answer 1

NaCN + H2O <----> Na+ OH- +H+ + CN- let x be the conc of OH; for each molecule of OH dissociated by NaCN, one molecule of NaCN is lost. If x is the concentration of OH-

[OH-] = x

Then the remaining concentration of NaCN is C - x, where C is the initial concentration (0.35M). The equilibrium constant is given by [OH-]*[Na+]/[NaCN], but this is a basic reaction so the equilibrium constant to use is Kb; You are given Ka. However, you should know that Kb*Ka = 10^-14, so Kb = 10^014/Ka = 1/4.9*10^-4

Using x for the concentration [OH-] and [Na+]

C - x ---> x + x

Kb = x^2/(C-x) X^2 = Kb*C - Kb*x x^2 + Kb*x - Kb*C = 0

x = 0.5*{-Kb ± √[kb^2 + 4*Kb*C]}

x = [OH-] 0.0027M

This concentration is small enough compared to 0.35M that you could use the expression x^2/C = Kb, x = √[Kb*C]and get the same result. So your only error was using Ka instead of Kb.

You get pOH = 2.58 and Ph = 11.42

Answer 2

I don't see what you're doing wrong...The only thing I can suggest should give you the same answer.

10^-14 / 4.9x10^-10 = Kb for NaCN

Go through the same calculations as you did for HCN.

Say that .35 is present initially, x forms, so x^2 / .35 - x = whatever.

Then that number would be the [OH].

The only other thing that I could see that would change your answer(to a very, very, very small degree) would be the fact that you didn't make it a quadratic. That really shouldn't matter though.

Keep trying. I don't see anything wrong with how you went about doing the problem.