Calculate the pH at 25 degree Celsius of a 0.19M solution of potassium cyanide (KCN). Note that hydrocyanic acid (HCN) is a weak acid with a pKa of 9.21. Round your answer to 1 decimal place.

Question

Chemistry

Raiden Hendrix

9 February, 00:02

Calculate the pH at 25 degree Celsius of a 0.19M solution of potassium cyanide (KCN). Note that hydrocyanic acid (HCN) is a weak acid with a pKa of 9.21. Round your answer to 1 decimal place.

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Answers (1)

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Kathleen Cannon · Answer 1

Answer is: pH of solution of potassium cyanide is 11.2.

Chemical reaction 1: KCN (aq) → CN⁻ (aq) + K⁺ (aq).

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

c (KCN) = c (CN⁻) = 0.19 M.

pKa (HCN) = 9.21.

Ka (HCN) = 10∧ (-9.21) = 6.16·10⁻¹°.

Kb (CN⁻) = 10⁻¹⁴ : 6.16·10⁻¹° = 1.62·10⁻⁵.

Kb = [HCN] · [OH⁻] / [CN⁻ ].

[HCN] · [OH⁻] = x.

[CN⁻] = 0.19 M - x ...

1.62·10⁻⁵ = x² / (0.19 M - x).

Solve quadratic equation: x = [OH⁻] = 0.00174 M.

pOH = - log (0.00174 M) = 2.76.

pH = 14 - 2.76 = 11.2.