I have 5 questions left on my review?

Hello, I am studying a review sheet for a quiz and there are five questions i cant seem to answer. Hope u can help….thanx!

Determine the Ka of an acid whose 0.294 M solution has a pH of 2.80

Lets call the acid HA, then HA<==>H+ + A-
Ka=[H+][A-]/[HA]
We have [HA]=0.294 and since we have pH we can find [H+] and [A-]
[H+]=[A-]=10^-2.80=0.0016
Ka=(0.0016)²/0.294
=8.7×10^-6
d) 8.6×10^-6

Determine the pH of a 0.188 M NH3 solution. The Kb of NH3 is 1.76×10^-5.

NH3 + H2O<==>NH4+ + OH-
Kb=[NH4+][OH-]/[NH3] (remember H2O isn’t included because it is liquid)
1.76×10^-5=x²/(0.188-x)
You can ignore the x in the denominator because 0.188-x is almost the same as 0.188.
1.76×10^-5=x²/0.188
x=[OH-]=0.00182
pH=14-pOH=14+log(0.00182)=11.260
a) 11.260

Which of the following compounds solubility will not be affected by a low pH solution?
This has to do with le chatelier’s principle. If you look at CuS for example. In water the equation is CuS<==>Cu2+ + S2-
Now low pH means acidic meaning H+ concentration is greater than OH-. So if you add acid to CuS solution then the H+ ions react with S2- to form HS- because HS- is a weak acid. Now for AgCl you have AgCl<==>Ag+ + Cl-, if you add H+ nothing happens because H+ doesn’t react with Cl- to form HCl because HCl is a strong acid, so AgCl would not be affected by a low pH since Cl- doesn’t react with H+.
a) AgCl

Which of the following statements is true
if Q>K then you have too many products and the reaction will proceed left to form reactants. If Q c) if Q=K, it means the reaction is at equilibrium

Determine the Kb for CN^-. The Ka for HCN is 4.9 x 10^-10.
Kw=1.0x10^-14=Ka*Kb=4.9x10^-10*Kb
Kb=2.0x10^-5
e) 2.0 x10^-5

Bath Product Review: PHILOSOPHY EGGNOG