Which response has both answers correct? Will a precipitate form when 250 mL of

Question

Which response has both answers correct?   

Will a precipitate form when 250 mL of 0.33 M Na2CrO4 are added to 250 mL of 0.12 M AgNO3? [Ksp(Ag2CrO4) = 1.1 × 10–12] What is the concentration of the silver ion remaining in solution?

A)       Yes, [Ag+] = 2.9 × 10–6 M.                    D)       No, [Ag+] = 0.060 M.

B)       Yes, [Ag+] = 0.060 M.                           E)       No, [Ag+] = 0.105 M.

C)    Yes, [Ag+] = 1.3 × 10–4 M.

Part 1:

Ksp = [Ag+]^2[CRO4]

Q = [0.06 M]^2[0.165 M]

Q = 5.9 x 10^-4

Q > Ksp

Yes, a precipitate will form.

Part 2:

[Ag2CRO4] = 2[Ag+][CRO4]

-s +2s +s

1.1 x 10^-12 = 4s^3

s = 6.5 x 10^-5

Ag+ ion concentration = 1.3 x 10^-4

That is as far as I get. I just can't seem to grasp how to get the remaining ion concentration. Any help would be appreciated. Thanks.

Lastly. the ANSWER is A. I just don't know how they got it. Thanks.

Explanation / Answer

Solution :-

Q> ksp means precipitate will form

After mixing the solutions the volume will be 250 ml +250 ml = 500 ml

Therefore the concentrations of the both compound will be halved

Therefore new concentrations are as follows

Na2CrO4 = 0.33 M /2 = 0.165 M

AgNO3 = 0.12 M / 2 = 0.06 M

Now lets calculate the concentration of the Ag+ remain after the precipitation

Ksp = [2Ag^+]^2 [ CrO4^2-]

1.1*10^-12 = [2x]^2 * [0.165]

1.1*10^-12 / 0.165 = 4x^2

6.66*10^-12 = 4x^2

6.66*10^-12/ 4 = x^2

1.66*10^-12 = x^2

Taking square root of both sides we get

1.29*10^-6 = x

Therefore the concentration of the Ag^+ = 1.29*10^-6 M

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