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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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Dejanras
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.

dsdrajlin

Answer:

11.3

Explanation:

KCN is a strong electrolyte that dissociates in water according to the following equation.

KCN(aq) → K⁺(aq) + CN⁻(aq)

The molar ratio of KCN to CN⁻ is 1:1. Then, [CN⁻] = 0.19 M.

CN⁻ is a base that reacts with water according to the following equation.

CN⁻(aq) + H₂O(l) = HCN(aq) + OH⁻(aq)

Given pKa of HCN, we can calculate pKb of CN⁻ using the following expression.

pKa + pKb = 14

pKb = 14 - pKa = 14 - 9.21 = 4.79

The basic dissociation constant (Kb) is:

pKb = -log Kb

Kb = antilog -pKb = 1.62 × 10⁻⁵

We can find the concentration of OH⁻ using the following expression.

[OH⁻] = √(Kb × Cb)

[OH⁻] = √(1.62 × 10⁻⁵ × 0.19) = 1.8 × 10⁻³ M

The pOH is:

pOH = -log [OH⁻]

pOH = -log 1.8 × 10⁻³ = 2.7

The pH is:

pH = 14 - pOH

pH = 14 - 2.7 = 11.3

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