Question

Calculate the value of the equilibrium constant, K c , for the reaction AgI ( s ) + 2 CN − ( aq ) − ⇀ ↽ − [ Ag ( CN ) 2 ] − ( aq ) + I − ( aq ) K c = ? The solubility product constant, K sp , for AgI is 8.52 × 10 − 17 , and the formation constant, K f ( β 2 ), for [ Ag ( CN ) 2 ] − is 1.0 × 10 21 .

Answer 1

Answer:

equilibrium constant Kc = 8.52 × 10⁻⁴

Explanation:

You  need The equations for the dissolution of AgI(s) and the stability constant of the  complex [Ag(CN)₂]⁻(aq)

The equations for the dissolution of AgI(s) is:

AgI(s)   ⇔   Ag⁺(aq)  +  I⁻(aq)               Ksp  =   [Ag⁺(aq) I⁻(aq)]

The stability constant of the  complex [Ag(CN)₂]⁻(aq) is:

Ag⁺(aq)  +     2CN⁻(aq)    ⇔  [Ag(CN)₂]⁻(aq)        Kfb2 = $\frac{[Ag(CN)_{2}(aq)]^{-} }{[Ag^{+}(aq)][2CN^{-}(aq)] }$

Adding the equations for the dissolution of AgI(s) and the stability constant of the  complex [Ag(CN)₂]⁻(aq) we get the required reaction. so the equilibrium   constant for the reaction is the product of the individual equilibrium constants:

AgI(s)     +     2CN⁻(aq)  ⇔   [Ag(CN)₂]⁻(aq)   +  I⁻(aq)

Kc = Ksp  × Kfb2  =    [Ag⁺(aq) I⁻(aq)]   ×  $\frac{[Ag(CN)_{2}(aq)]^{-} }{[Ag^{+}(aq)][2CN^{-}(aq)] }$  =  $\frac{[Ag(CN)_{2}(aq)]^{-} [I(aq)^{-}] }{[2CN^{-}(aq)] }$

Kc =  8.52 × 10⁻¹⁷  ×  1  × 10²¹  = 8.52 × 10⁻⁴