Determine the molar solubility of MX (Ksp=4.2x10-8) in 0.047 M NaCN. The metal ion M+ forms M(CN)2- (Kf=3.1x1010) in the presence of CN- ion
..................MX ==> M^+ + X^- and Ksp = (M^+)(X^-) = 4.2E-8
I................solid.........0.........0
C...............solid.........S.........S where S = solubility
But M^+ + 2CN^- ==> [M(CN)2]^- and Kf = 3.1E10
3,1E10 = [M(CN)2]^-/(M^+)(CN^-)^2
You know total solubility, S = (M^+)[M(CN)2]-
Use Kf to solve for ratio of [M(CN)2]^-/(M^+),plug in CN^- and solve for [M(CN)2]- in terms of (M^+), plug that in to the total S equation above and finally solve for (M^+) in terms of S. Plug that into Ksp expxression and solve for S. Post your work if you get stuck.
To determine the molar solubility of MX in 0.047 M NaCN solution, we need to consider the formation of the complex ion M(CN)2-.
Let's assume the initial concentration of MX is "x" mol/L.
The reaction between MX and NaCN can be represented as follows:
MX (s) + 2 NaCN (aq) ⇌ M(CN)2- (aq) + 2 NaX (s)
Using the given equilibrium constant (Kf) for the formation of M(CN)2-, we can write the expression for the formation constant:
Kf = [M(CN)2-] / [MX] * [CN-]^2
Since MX is a solid, its concentration remains constant and can be treated as 1 for simplicity.
Therefore, Kf = [M(CN)2-] / [CN-]^2
Given that Kf = 3.1x10^10 and the concentration of NaCN is 0.047 M, we can substitute these values into the equation and rearrange it to solve for [M(CN)2-].
3.1x10^10 = [M(CN)2-] / (0.047)^2
[M(CN)2-] = 3.1x10^10 * (0.047)^2
[M(CN)2-] = 6.938x10^6
Now, we need to convert the concentration of M(CN)2- to molar solubility of MX.
Since the stoichiometry of the reaction is:
1 mol MX produces 1 mol M(CN)2-
The molar solubility of MX is equal to the concentration of M(CN)2-:
Molar solubility of MX = [M(CN)2-] = 6.938x10^6 mol/L
Therefore, the molar solubility of MX in 0.047 M NaCN solution is 6.938x10^6 mol/L.
To determine the molar solubility of MX in 0.047 M NaCN, we need to consider the solubility product constant (Ksp) and the formation constant (Kf).
1. Write the balanced chemical equation for the dissolution of MX:
MX ⇌ M+ + X-
2. Write the balanced equation for the formation of M(CN)2-:
M+ + 2CN- ⇌ M(CN)2-
3. Set up an ICE (Initial, Change, Equilibrium) table for the dissolution of MX:
MX ⇌ M+ + X-
Initial: 0 0 0
Change: +s +s +s
Equilibrium: s s s
Here, 's' represents the molar solubility of MX.
4. Set up an ICE table for the formation of M(CN)2-:
M+ + 2CN- ⇌ M(CN)2-
Initial: s 2s 0
Change: -s -2s +s
Equilibrium: s-s 2s-2s s
Note that the concentration of CN- in solution is given as 0.047 M.
5. Write the equilibrium expression for the dissolution of MX:
Ksp = [M+][X-] = s*s = s^2
6. Write the equilibrium expression for the formation of M(CN)2-:
Kf = [M(CN)2-]/[M+][CN-]^2 = s/(s^2 * (0.047)^2)
7. Since M(CN)2- is formed from M+ and CN-, we can use the formation constant to express the concentration of M+ in terms of s:
[M+] = [M(CN)2-]/(2[CN-]) = s/(2*0.047)
8. Substitute the values into the equilibrium expression for the formation of M(CN)2-:
Kf = [M(CN)2-]/[M+][CN-]^2 = (s/(s^2 * (0.047)^2))/((s/(2*0.047))*(0.047)^2)
= 3.1x10^10
9. Simplify the equation:
s/(s^2 * (0.047)^2) * ((2*0.047)/(0.047)^2) = 3.1x10^10
2/(s * 0.047) = 3.1x10^10
2 = 3.1x10^10 * s * 0.047
s = 2 / (3.1x10^10 * 0.047)
10. Calculate the value of s:
s ≈ 1.3x10^-11 M
Therefore, the molar solubility of MX in 0.047 M NaCN is approximately 1.3x10^-11 M.