Question

Determine the pH of an aqueous solution of Fe(CN)₆^3-, prepared using FeCl₃ and NaCN with formal concentrations of 0.001832 M and 0.5120 M.

logK_f(Fe(CN)₆^3-) = 43.6 K_a(HCN) = 6.2x10^-10

Answer 1

Given that; logK_f(Fe(CN)₆^3-) = 43.6

Kf = 4.0*10^43

Kf = = |Fe(CN)63-| / (|Fe3+| * |CN-|^6)

|[Fe(CN)6]3-| = |[Fe(CN)6]---|° - (|CN-| / 6)
|Fe3+| = |CN-| / 6

Put this value in the following equation:

Kf = = |[Fe(CN)6]---| / (|Fe3+| * |CN-|^6)

The Equilibrium of formation is as follows:

Kf = 4.0 *10^43

Kf = [|[Fe(CN)6]---|° - (|CN-| / 6)] / |CN-|^7

To solven this we will get

|CN-| = 4.5*10^-7 M

The CN- reacts with water as follows:

CN-(aq) + H2O(aq) <---> HCN(aq) + OH-(aq)

The equilibrium Constant for the above equation:

K,hydr = Kw / Ka,HCN

Here K_a(HCN) = 6.2x10^-10

Kw = 1.0*10^-14

K,hydr = 1.0E-14 / 6.2E-10

= 1.6*10^-5
K,hydr =|HCN| * |OH-| / |CN-|

1.6*10^-5 =|HCN| * |OH-| / |CN-|

here |CN-| = 4.5*10^-7 M and |HCN| =|OH-|

1.6*10^-5 =|OH-|^2 / (4.5*10^ -7 - |OH-|)

To solve this we will get:

|OH-| = 4.3*^-7 M
so pOH = -log |OH-| = -log |4.3*^-7 M|

pOH =6.4

and pH+pOH = 14

pH = 14.0-6.4

pH =7.6