# How do we calculate the molar solubility of basic salts at different pH's?

been working on this problem all day and it doesn't make any sense to me

"find the molar solubility of calcium hydroxide in a solution buffered at each of the following pH's"

a. ph=4

b. ph=7

c. ph=9

Here's what I did for the first part

ph=4 , pOH=10

[OH] = 1.0E-10

Ca(OH)2 ---> Ca+2 + 2OH-
I - 1.0E-10M
C +x 1.0E-10M+2x
E x (1.0E-10M + 2x)^2

Ksp = 4.68E-6

I can plug in these values directly and wont get a reasonable answer for x

I have also figured that since the Ksp is orders of magnitude greater than the initial hydroxide concentration we can ignore it and thus assume its just the same as aqueous

wasted 5 hours of my life on this... hate myself

D

#### deleted936470

ICE? ew. don't.
Ksp @ 25°C = 5.02×10^–6 & molar solubility (s) of Ca(OH)2*[OH⁻] = 10^(pH - 14)
Ksp = s * (10^(pH - 14))^2 = s * 10^(2*pH - 28)
s = Ksp / 10^(2*pH - 28) = Ksp * 10^(28 - 2*pH)
for example, at neutral pH
s = 5.02*e–6*10^(28 - 2*7) = 5.02*e–6*10^(14) = 5.02*e8 mol/L

#### PlsLetMeIn21

2+ Year Member
It's all in the bigger picture with your question.

Ksp = [Ca2+][OH-]^2, so [Ca2+] = Ksp/[OH-]^2 where [Ca2+] is your molar solubility.

a) molar solubility = Ksp/(10^-10)^2 = Ksp x 10^20 = 5 x 10^-6 x 10^20 = 5 x 10^14

b) molar solubility = Ksp/(10^-7)^2 = Ksp x 10^14 = 5 x 10^-6 x 10^14 = 5 x 10^8

c) molar solubility = Ksp/(10^-5)^2 = Ksp x 10^10 = 5 x 10^-6 x 10^10 = 5 x 10^4

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