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.

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

12 is it do you think so