What is the ph of a 0.25 m solution of nacn? (the pka of hcn is 9.21.) (a) 4.91 (b) 8.61 (c) 11.30 (d) 13.40?

Answer is: pH of solution of sodium cyanide is (c) 11.30.

Chemical reaction 1: NaCN(aq) → CN⁻(aq) + Na⁺(aq).

Chemical reaction 2: CN⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻) = 0.25 M.

pKa(HCN) = 9.21.

Ka(HCN) = 10∧(-9.21).
Ka(HCN) =  6.16·10⁻¹⁰.

Ka(HCN) · Kb(CN⁻) = 10⁻¹⁴.
Kb(CN⁻) = 10⁻¹⁴ ÷ 6.16·10⁻¹⁰ 

Kb(CN⁻) = 1.62·10⁻⁵.
Kb = [HCN] · [OH⁻] / [CN⁻].
[HCN] · [OH⁻] = x.
[CN⁻] = 0.25 M - x..
1.62·10⁻⁵ = x² / (0.25 M - x).
Solve quadratic equation: x = [OH⁻] = 0.002 M.
pOH = -log(0.002 M) = 2.70.
pH = 14 - 2.70 = 11.30.

answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e