chemistry

Hi,
I was hoping someone could double check a chemistry answer for me. I am trying to calculate the Ksp of Ag2CrO4. This question is based on a experiment to determine ksp experimentally. In the experiment the Ksp of silver chromate was calculated using a single displacement reaction between a saturated solution of silver chromate and copper metal. A 1.00 L of saturated silver chromate was used in the experiment. Before the reaction the mass of copper was 1.250 g and after it was 1.240 g. In moles this was calculated to be 0.00016 mol of Cu. In the balanced equation, the copper and silver chromate reactants have a mole ration of 1.1. The balanced reaction is Cu + Ag^2CrO^4 ----> CuCrO^4 + 2Ag. The Ksp value measured for silver chromate was given as 1.1 x 10^-12.

So given that information here is what I did.

Ksp = [Ag^+]^2 [CrO4^2-]
The molar mass of Ag^2CrO4 is 331.73 g/mol

0.00016/ 331.73 = 4.8 x 10^-7

(4.8 x 10^-7)^2(4.8 x 10^-7) = 1.10 x 10^-19

Then I calculated my percentage error.

1.10 x 10^-19 - 1.1 x 10^-12/ 1.1 x 10^-12 x 100% = 9.9 x 10^-23

Did I solve it correctly? If not could someone show me where I went wrong. Thank you in advance.

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asked by Laura
  1. I would rethink the problem.
    1.6E-4/molar mass Ag2C4O4 give I don't know what.
    Cu + Ag2CrO4 ==> 2Ag + CuCrO4
    If you had 1.6E-4 mols Cu you must have had 1.6E-4 mols Ag2CrO4.
    CrO4^2- = 1.6E-4
    Ag^+ = 2*1.6E-4
    Calculate Ksp from that.
    Your % error, from your calculation is 100% because of a math error (but recalculated from above it should be better than that but I don't know how much better).

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  2. 57. Determining K_sp experimentally
    a. Calculate K_sp of Ag_2 CrO_4. Organize your answer so that your solution is logical. Make sure that your K_sp has correct significant figures.
    a. From the experiment CU = 0.00016 mol. Also 〖Ag〗_2 〖CrO〗_4 = 0.0016 mol
    〖Ag〗_2 〖CrO〗_4⇋〖2Ag〗_((aq))^++ CrO_(4 (aq))^(2-) K_(sp )=1.1*〖10〗^(-12)
    [〖Ag〗_2 〖CrO〗_4 ]=n/V=0.00016mol/1.00L=1.6*〖10〗^(-4) mol/L

    ICE [Ag^+ ]^2 mol/L [〖CrO〗_4^(2-) ] mol/L

    Initial 0 0
    Change +2x +x
    Equilibrium 2(0.00016) = 0.00032 0.0016

    K_sp=[Ag^+ ]^2 [〖CrO〗_4^(2-) ]
    1.1*〖10〗^(-12)=(3.2*〖10〗^(-4) )^2 (1.6*〖10〗^(-4) )
    =1.6*〖10〗^(-11)
    b. Calculate percent error
    % error= ((experimental-Theoretical))/Theoretical*100
    =((1.6*〖10〗^(-11)-1.1*〖10〗^(-12)))/(1.1*〖10〗^(-12) )*100
    = 13.89% = 14%

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    posted by Alwiya

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