You just need to get started.
A2B3 ==> 2A^+3 + 3B^-2
If pKsp = 82.55, then take the antilog of both sides which is
antilog pKsp = antilog 82.55
Ksp = 3.548 x 10^82
(Do you know how to do that. Punch in 82.55 on your calculator, then hit the 10^x key and that will give you the antilog 82.55 which is 3.548 x 10^82.)
I suspect you can take it from here. If not, tell us what you don't understand and show your work to the point at which you don't know what to do next.
Ksp = (A)^2(B)^3 = 3.548 x 10^82.
Then set and solve for (A2B3).
Consider a salt of the composition, A2B3. For this salt, pKsp=82.55. What is the logarithm of the molar concentrtion of this salt?
I'm really at a loss of where to begin. We didn't have any examples like this in class. Thanks!
4 answers
step 1:[2A^+3][3B^-2]/[A2B3]=82.55=Ksp
step 2:total concentration of the top part is [3.548E82]/x=82.55
step 3:[3.548E82]x 82.55=x
step 4: x=2.92E84
Can you please check this if am correct and am in the right path...am just not good with this chemisty stuff...forgive my incorrect assumpions if any.
step 2:total concentration of the top part is [3.548E82]/x=82.55
step 3:[3.548E82]x 82.55=x
step 4: x=2.92E84
Can you please check this if am correct and am in the right path...am just not good with this chemisty stuff...forgive my incorrect assumpions if any.
I wouldn't do it that way.
A2B3 ==> 2A^+2 + 3B^-3
Ksp = (A)^2(B)^3
Let X = (A2B3) which is what you want to solve for.
If X = (A2B3), then (A^+2) = 2X and (b^-3) = 3X. Now substitute that into the Ksp expression.
Ksp = (2X)^2((3X)^3 = 3.548E82
Now solve for X which is the molar concentration of (A2B3).
Then the problem asks you to take the log of that
A2B3 ==> 2A^+2 + 3B^-3
Ksp = (A)^2(B)^3
Let X = (A2B3) which is what you want to solve for.
If X = (A2B3), then (A^+2) = 2X and (b^-3) = 3X. Now substitute that into the Ksp expression.
Ksp = (2X)^2((3X)^3 = 3.548E82
Now solve for X which is the molar concentration of (A2B3).
Then the problem asks you to take the log of that
Assuming the final concentration of chloride anion after the addition of HCl (precipitation step) was 0.1M, what is the remaining concentration of Ag+ in the solution? (pKsp for AgCl = 9.74)