Question

1A.If 14.0 mL of 0.10 M Ba(NO₃)₂ are added to 48.0 mL of 0.17 M Na₂CO₃, will BaCO₃ precipitate?

A.BaCO₃ will precipitate.

B. BaCO₃ will not precipitate.     

Calculate Q. ______________

1B The molar solubility of MnCO₃ is 4.2 ? 10^-6M. What is K_sp for this compound?

Answer 1

V = 14ml

M = 0.1 Ba(NO3)2

V2 = 48 ml

M2 = 0.17 Na2CO3

If Q > Ksp, then it will precipitate

Ba+2 and CO3-2 ---> BaCO3

Ksp = [Ba+2][CO3-] = 5.1*10^-9

[Ba+2] = 0.1

[CO3-2) = 0.17

Q = [Ba+2][CO3-] = 0.1*.17 = 0.017

Since Q>Ksp, expect precipitate... choose A

FOR

1B

M = 4.2*10^-6 MnCO3

MnCO3 ---> Mn+2 and CO3-2

Ksp = [Mn+2][CO3-2]

Since [Mn+2] = [CO3-2] let us assign it the variable "s" beacuase 1 mol of MnCO3 will be soluble for each

Ksp = s*s

Ksp = (4.2*10^-6)^2

Ksp = 1.76*10^-11