Calculate the ph of a 0.021 m nacn solution. [ka (hcn) = 4.9  10-10]

Question

Chemistry

Willie Casey

2 April, 08:04

Calculate the ph of a 0.021 m nacn solution. [ka (hcn) = 4.9  10-10]

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Answers (1)

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Melody Edwards · Answer 1

Answer is: pH of solution of sodium cyanide is 11.3.

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

Chemical reaction 2: CN ⁻ + H₂O (l) ⇄ HCN (aq) + OH⁻ (aq).

c (NaCN) = c (CN ⁻) = 0.021 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 ⁻ ] = 0.021 M - x ...

2.04·10 ⁻⁵ = x² / (0.021 M - x).

Solve quadratic equation: x = [OH ⁻ ] = 0.00198 M.

pOH = - log (0.00198 M) = 2.70.

pH = 14 - 2.70 = 11.3.