Question

1. Determine the pH of a solution containing 1.00 mol/L of HCN (Ka = 5.00×10–10) and 5.00×10–2 mol/L of HCl.

Answer 1

Essentially one ignores the contribution of H^+ from the HCN but you can calculate how much that is.
.......HCN ==> H^+ + CN^-
I......1........0.....0
C......-x.......x.....x
E......1-x......x.....x

For HCl, remember it is a strong electrolyte and ionizes completely.
........HCl ==> H^+ + Cl^-
I.....0.05......0......0
C....-0.05....0.05....0.05
E........0....0.05....0.05
E....

Ka for HCN = (H^+)(CN^-)/(HCN)
(H^+) = 0.05+x (that's 0.05 from the HCl and x from the HCN)
HCN is 1-x. Solve for (CN^-) and that = (H^+) from the HCN.
Compare with that from the HCl. You will see it is negligible.
Then pH = -log(HCl) = ?

Answer 2

P.S. Note the correct spelling of chemistry.