Question

Calculate the Ksp of Ag2CrO4.

Answer 1

I'm not sure if i'm right or not.

The mass of copper before reaction: 1.250 g
The mass of copper after reaction: 1.240 g
The mass of copper: (1.250-1.240=0.010g)
Moles: 0.010/63.5=0.000157 mol of Cu

Balanced equation: Cu(s) + Ag (2)CrO4(aq) ----> CuCrO4(aq) + 2Ag(s)

The concentration of a saturated solution of silver chromate is 0.00015748mol/L
[Ag2CrO4(aq) ] =n/v=0.00015748/1L=1.57x10^-4

Ag2CrO4(s) ⇌ 2Ag(aq)^+ + CrO4^2-
Ksp=[Ag+]^2[CrO4^2-]

ICE [Ag^+] in mol/L [CrO4^2-] in mol/L
Initial 0 0
Change +2x(1.57x10^-4) +x(1.57x10^-4)
Equilibrium 3.14x10^-4 1.57x10^-4

Ksp=[Ag+]^2[CrO4^2-]
Ksp = [3.14x10^-4]^2[1.57×10^-4]
Ksp=1.55x10^-11

Therefore the Ksp is 1.55x10^-11

b)
% error=(experimental-theoretical)/theoretical x100%
=1.55x10^-11 - 1.1x10^-12/1.1x10^-12x100%
=3.3x10^-23

or this is right.

Cu(s) + Ag2CrO4(aq) → CuCrO4(aq) + 2Ag(s)
mol of Cu ≡ 2 mol of Ag ≡ mol of [CrO4]^-
mol of Cu = 0.01/63.5 = 1.57×10^-4 mol (At Wt Cu= 63.5 no need for extra sig figs)
mol of Ag = ½[1.57×10^-4] mol
Ag2CrO4(s) ⇋ 2Ag^+(aq) + [CrO4]^2-(aq)
Ksp = [Ag^+]^2 [[CrO4]^2-]
Ksp = [(2×½) 1.57×10^-4]^2[1.57×10^-4]
Ksp = [1.57×10^-4]^3 = 3.9 ×10^-12 (two sig figs generous here should be 4 ×10^-12 given mass of Cu to only one sig fig)
% error = (3.9 ×10^-12- 1.1×10^-12)/1.1×10^-12 ≈ 250%

Errors: You should be weighing to three or four decimal places (analytical balance)! Your loss is 0.01± 0.01??
How did you make sure all the Ag was removed from the Cu before weighing?
Was copper lost at the removal of Ag stage? Was the copper dried after removing the Ag?