Question
(Ksp= 5.0*10^-13)
1) Calculate the molar solubility of AgBr in 3.0×10^−2 M AgNO3 solution.
2) calculate the molar solubility of AgBr in 0.10 M NaBr solution.
I tried solving using this the quadratic formula, but it didn't work.
I started with these equations and then solved for x, but none of them gave me the correct answer.
1) 5.0 x 10^-13 = (x + 3.0 x 10^-2)(x)
2) 5.0 x 10^-13 = (x)(0.10+x)
1) Calculate the molar solubility of AgBr in 3.0×10^−2 M AgNO3 solution.
2) calculate the molar solubility of AgBr in 0.10 M NaBr solution.
I tried solving using this the quadratic formula, but it didn't work.
I started with these equations and then solved for x, but none of them gave me the correct answer.
1) 5.0 x 10^-13 = (x + 3.0 x 10^-2)(x)
2) 5.0 x 10^-13 = (x)(0.10+x)
Answers
THANKS for showing your work. If you had shown your math work I could have checked the error. You must have made a math error. The chemistry is OK. And you need not solve the quadratic with either of them.
How would I do it without the quadratic formula?
Make the assumption that x is small in comparison to 0.03(another way of saying x + 0.03 = 0.030). Then
5.0E-13 = (x)(0.03)
x = (5.0E-13/0.03) = 1.67E-11
Then you check the assumption to see if you must go back and solve the quadratic. So 1.67E-11 + 0.03 = essentially 0.03---not even close). Since the Ksp has only 2 s.f. in it I would round that 1.67E-11M to 1.7E-11M.
You were right to include the x in the set up because some problems will not allow you to ignore the x. I always make the assumption, then check it when I finish. If the easy answer is within 5% (I think most texts use the 5% rule now) I let it stand. If the error is more than that I go back and solve the quadratic.
5.0E-13 = (x)(0.03)
x = (5.0E-13/0.03) = 1.67E-11
Then you check the assumption to see if you must go back and solve the quadratic. So 1.67E-11 + 0.03 = essentially 0.03---not even close). Since the Ksp has only 2 s.f. in it I would round that 1.67E-11M to 1.7E-11M.
You were right to include the x in the set up because some problems will not allow you to ignore the x. I always make the assumption, then check it when I finish. If the easy answer is within 5% (I think most texts use the 5% rule now) I let it stand. If the error is more than that I go back and solve the quadratic.