Question

The formation constant* of [M(CN)2]– is 5.30 × 1018, where M is a generic metal.

A 0.160-mole quantity of M(NO3) is added to a liter of 0.720 M NaCN solution. What is the concentration of M ions at equilibrium?

Answer 1

The formation of the compound can be described as

M⁺ + 2 CN^- ========= [M (CN)₂]^-

We know that we have a volume of 1 liter then the initial concentrations are

M = 0.16 M

CN = 0.72 M

the next thing to do is the ICE chart

............M⁺ ......................+ 2 CN^- ========= [M (CN)₂]^-

I.......... 0.16....................... 0.72 .............................0

C.......... -x......................... -2x.................................. +x

E ..........0.16-x.................. 0.72-2x ..............................x

we know we have a kf of 5.3 x 10¹⁸ , this is a very big value, we can assume that almost all the M⁺ ions will be consumed since this is the limiting reactant, we can apply a simplification, the ICE chart will be like this:

.................M⁺ + .........................2 CN^- ========= [M (CN)₂]^-

I................ 0.16 ..........................0.72 ...........................0

C ...............-0.16...................... -2(0.16) ........................0.16

E.................. y............................. 0.4............................. 0.16

y is the ammount of M+ ions at the end, it must be a very small value since the K formation is too big so remember that

Equilibrium constant

aA + bB → cC + dD

Equilibrium constant is the relationship of products and reactants expressed like this

K = [A]^a [B]^b / ( [C]^c[D]^d )

K is the equilibrium constant, for our particular case

K =  [M (CN)₂]^- / ( [M⁺]*[CN]²)

5.3 x 10¹⁸ = 0.16 / (s*0.4²)

5.3 x 10¹⁸ = 0.16 / ( s * 0.16 )

5.3 x 10¹⁸ = 1 / s

s = 1.89 x 10^-19 M , this is the equilibrium concentration of M⁺ ions

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