Ka for hcn is 4.9 ⋅ 10-10. what is the ph of a 0.068 m aqueous solution of sodium cyanide? ka for hcn is 4.9 10-10. what is the ph of a 0.068 m aqueous solution of sodium cyanide? 0.74 7.00 2.96 13.24 11.07

Answer is: pH of solution of sodium cyanide is 11.07.
Chemical reaction 1: NaCN(aq) → CN⁻(aq) + Na⁺(aq).
Chemical reaction 2: CN⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻) = 0.068 M.
Ka(HCN) =  4.9·10⁻¹⁰.
Kb(CN⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵.
Kb = [HCN] · [OH⁻] / [CN⁻].
[HCN] · [OH⁻] = x.
[CN⁻] = 0.068 M - x..
2.04·10⁻⁵ = x² / (0.068 M - x).
Solve quadratic equation: x = [OH⁻] = 0.00116 M.
pOH = -log(0.00116 M) = 2.93.
pH = 14 - 2.93 = 11.07.

43.452

step-by-step explanation:

284 times .153 = 43.452