Acids / Bases - Need some clarity for final exam I'm studying for my final exam, and these are the questions I got wrong on the mid term, could anyone help correct me? I know people are against giving answers for homework, so I must add that is now exam period for Universities, so this is only for my further understanding!
Thanks so much! 



My attempts will be in this format



1)The pH of a 0.2M unknown base X OH is 8.15

a)Calculate [OH-]



10^(-8.15) = 7E-9M



b)What is the % disassociation



(7E-9/0.2M) x 100 = 3.5E-6%



c)What is the Kb of the base?



This one I don't even know where to begin..



2)Calculate the pH of the following solutions

a) 0.10M KCN, Ka (HCN) = 6.2E-10 (Hint: KCN is the salt of a weak acid and a strong base)



x = sqrt(Ka*c)
x = sqrt(6.2E-10 * 0.1) = 7.8E-6
-log(7.8E-6) = pH = 5.1
1a) First use $pOH = 14 - pH$

Then use: $pOH = \log([OH^-])$

b) Simply use: $\dfrac{\text{dissociated}\ [OH^-]}{\text{original}\ [OH^-]}$

c) Use $K_b = \dfrac{[X^+][OH^-]}{[XOH]}$

2) Dissolving $KCN$ in water gives: 
$\ce{KCN \rightarrow K^+ + CN^-}$

The following equilibrium reaction is thus formed

$\ce{CN^- + H2O \rightleftharpoons HCN + OH^-}$

Which is:

$K_b = \dfrac{[HCN][OH^-]}{[CN^-]}$

From the $K_a$, you can get the $K_b$ through $K_w\ =\ K_aK_b$ so that:

$\dfrac{K_w}{K_a} = \dfrac{[HCN][OH^-]}{[CN^-]}$

Let the amount of $CN^-$ that reacts to be $x\ M$, thus $[CN^-]$ becomes $0.1-x$ and you get $x\ M$ of both $[HCN]$ and $[OH^-]$. Solve for $x$ to get $[OH^-]$ and from there, $pH$ should be easy to calculate.

3) Use $Ksp = [Ba^{2+}][F^-]^2 = 1.7\times 10^{-6}$ and solve for $[Ba^{2+}]$.