Question

In: Chemistry

Use systematic equilibria calculation to find the concentrations of Ag+, CN-, and HCN in a saturated solution of AgCN whose pH is fixed at 9.00

Use systematic equilibria calculation to find the concentrations of Ag+, CN-, and HCN in a saturated solution of AgCN whose pH is fixed at 9.00. Consider the following equilibria. AgCN(s)  Ag+ + CN- Ksp = 2.2 x 10-16 CN- + H2O  HCN (aq) + OH- Kb = 1.6 x 10-5

Expert Solution

Given,

AgCN(s) ---> Ag⁺ + CN^- Ksp = 2.2 x 10^-16

Ksp = [Ag⁺] [CN^-]

2.2 * 10^-16 = x²

x = 1.48 * 10^-8

[Ag⁺] = 1.48 * 10^-8 and [CN^-] = 1.48 * 10^-8

As we know, pH = -log[H⁺] and [H⁺] = antilog(-pH)

[H⁺] = antilog(-9) = 0.0001M

pH + pOH = 14

SO, pOH= 14-9= 5

pOH = -log[OH^-] and [OH^-] = antilog(-pOH)
[OH^-] = antilog(-5) = 0.0067M

Hence, [H₂O] = 0.0001M + 0.0067M = 0.0068M (Because H₂O ----> H⁺ + OH^- ]

CN^- + H₂O ----> HCN (aq) + OH^- (Kb = 1.6 x 10^-5)

Kb = [HCN] [OH^-] / [CN^-] [H₂O]

1.6 x 10^-5 = [HCN] 0.0067M / (1.48*10^-8 M) (0.0068M)

[HCN] = 2.4 * 10^-13M

queen_honey_blossom answered 2 years ago

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