7. For each of the following situations, determine whether or not a precipitate

Question

7. For each of the following situations, determine whether or not a precipitate of MgF2 is expected to form. a) 500.0 mL of 0.050M Mg(NO3)2 is mixed with 500.0 mL of 0.010 M NaF. b) 500.0 mL of 0.050M Mg(NO3)2 is mixed with 500.0 mL of 0.0010 M NaF. 8. Is a precipitate of Cd(CN)2 expected to form when 500.0 mL of 0.010 M Cd(N03)2 is mixed with 500.0 mL of 0.0025 M NaCN? Both cadmium(II) nitrate and sodium(I) cyanide are completely dissociated in the original solutions. The Ksp of Cd(CN)2 is 1.0 X 10^-8.

Explanation / Answer

MgF2 --> Mg+2 and 2F-

Ksp =  7.4* 10^-11

Q = [Mg+2][F-]^2

a)

V = 500 ml of M = 0.05 Mg(NO3)2

V = 500 ml of M = 0.01 M of NaF

VT = 500´500 = 1000 ml or 1 L

[Mg+] = mol of MG+ / V = (M1*V1)/(1L) = (0.05*0.5)/1 = 0.25 M of MG+2

[F-] = mol of F- /V = (M2V2)/1L = 0.5*0.01/1 = 0.005 M of F-

Q = [Mg+2][F-]^2

Q = (0.25)(0.005)^2 = 6.25*10^-6

Since Q >>> Ksp, the solution will be supersaturated, therefore expect precipitate

b)

Apply same logic, same volumes, therefore V = 1 L

of Mg+, concetration remains the same

[Mg+] = mol of MG+ / V = (M1*V1)/(1L) = (0.05*0.5)/1 = 0.25 M of MG+2

For F-, concnetration changes:

[F-] = mol of F- /V = (M3V3)/1L = 0.5*0.001/1 = 0.0005 M of F-

Substitut ein Q

Q = [Mg+2][F-]^2

Q = (0.25)(0.0005)^2 = 6.25*10^-8

Q is still >>> than Ksp, expect precipitate

NOTE:

consider posting question 8 in another set of Q&A. We are not allowed to answer to multiple questions in a single set of Q&A

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