Acids / Bases - Need some clarity for final exam
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 $[Ba^{2+}]$.