Answer to Question #87103 in General Chemistry for hailey pulliam

Question

Calculate the pH of a 0.22 M solution of hydrocyanic acid (HCN), which has a pKa of 9.31.

Answer 1

V = 1L.

 

Xmol  Xmol   Xmol

 HCN ⇄ H⁺ +CN^-

 1mol  1mol 1mol

pK_a = 9.31;   K_a = 10^-9.31 = 4.90×10^-10.

K_a = [H^+][CN^-]/[HCN] = X²/(0.22-X) = 4.90×10^-10.

X² = 4.90×10^-10(0.22-X)

X²+4.9×10^-10×X–1.08×10^-10=0

D = b² – 4ac = (4.9×10^-10)² – 4×1×(–1.08×10^-10) = 4.32×10^-10.

$X_1=(-b-√D)/2a$;

X₁ = (–4.9×10^-10– √4.32×10^-10)/2 = –1.04×10^-5.

$X_2=(-b+√D)/2a$;

X₂ = (–4.9×10^-10+√4.32×10^-10)/2 = 1.04×10^-5.

X₁ does not match the condition of the question.

[H⁺] = X₂ = 1.04×10^-5.

pH = -lg[H⁺] = -lg(1.04×10^-5) = 4.9833 ≈ 4.98.