2 simple mastering chem questions :) all help appreciated

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Apr 29th, 2015

Part A: The answer is Q = 5.00 * 10^-6 M

Q, the ion product for the solution, is the product of the concentrations of the two ions in the solution. To find Q, you just need to multiply the two concentrations (for magnesium and carbonate) in the solution:

Q = (2.50 *10^-3)*(2.00*10^-3)

Q = 5.00*10^-6 M

Part B: The concentration of lead (Pb^2+) that must be exceeded is 4.67 * 10^-2 M.

The formation of PbBr_2 happens through the following reaction:

Pb^(2+) + 2Br^(-1) --> PbBr_2

The expression for Q is still the product of the ion concentrations, but since there need to be 2 bromide ions for every one lead ion, you need to square the concentration of bromide. To calculate Q for this reaction, the expression is:

Q = [Pb^2+][Br^-1]^2

You're looking to find the concentration of lead needed to precipitate PbBr_2, so you're looking for the concentration that makes Q greater than the value of Ksp, which is 4.67 * 10^-6. We also have a known concentration of bromide in this solution: 1.00*10^-2. Putting this all together, we get:

Q > 4.67 * 10^-6

[Pb^2+][Br^-1]^2 > 4.67 * 10^-6

[Pb^2+][1.00*10^-2]^2 > 4.67 * 10^-6

[Pb^2+](1.00*10^-4) > 4.67 * 10^-6

[Pb^2+] > ( 4.67 * 10^-6) / (1.00*10^-4) 

[Pb^2+] > 4.67 * 10^-2

Apr 29th, 2015

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