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.
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It might be best for whom? to give you the answer. We're trying to help you understand the problem. We aren't too interested in giving you the answer.
You have a strong acid (HCl) and a very weak acid (HCN). Usually the strong acid predominates and the pH is determined by the strong acid. You can calculate the total H^+ if you wish to see that the H^+ contributed by the HCN is negligible.
For HCl ==> H^+ + Cl^- and (H^+) from the problem is 0.05M.
For HCN --> H^+ + CN^-
I...1.0....0.05...0
C...-x.......x....x
E..1-x....0.05+x..x
NOtE: You MAY (or may not) need to solve the quadratic.
Ka = 5E-10 = (0.05+x)(x)/(1-x)
Solve this to find x; the total (H^+) then is 0.05+x. Then pH = -log(H^+). I think you will find that the HCN doesn't contribute much to the total H^+.