Answer to Question #143595 in General Chemistry for Johannes Steven

Question

Sodium cyanide is the salt of the weak acid HCN. Calculate the concentration of H3O^+ , OH^- , HCN and Na^+ in solution prepared by dissolving 10.8 g of NaCN in enough water to make 5.0 x 10^2 mL of solution at 25 degrees Celsius.

Answer 1

Solution:

Ka of HCN = 4.0×10^-10

M of NaCN = 10.8g NaCN/49.01g/mol NaCN/0.5L = 0.441 M

CN ^- + H₂O ↔ HCN + OH^-

0.441-x x x

Kb = 1×10^-14 / 4.0×10^-10 = 2.5×10^-5

Kb = [HCN] [OH-] / [CN-]

2.5×10^-5 = x²/(0.441-x) ≈ x²/0.441

x²=1.10×10^-5

x= 0.0033M

[HCN] = 0.0033M

[OH-] = 0.0033M

[Na+] = 0.44M

[H₃O⁺] = 3.0×10^-12M

Answer: [HCN] = 0.0033M; [OH^-] = 0.0033M; [Na⁺] = 0.44M; [H₃O⁺] = 3.0×10^-12M