How Do You Calculate Molar Solubility From Ksp?


Answer by Rudolf F. on Mar 3 2014

at equilibrium :
[Pb2+] = x
[CO32-]= x + 0.02

Ksp = 7.4 x 10^-14 = [Pb2+][CO32-] = (x)(x+0.02)

x = molar solubility = 3.7 x 10^-12 M

The Ksp of Cr(OH)2 is 6.7 E-31...help!?

Calculate its molar solubility in: a) neutral water b) in a solution buffered at pH 11 c) in a solution of .1M Cr(NO3)2 help! thanks.

Cr(OH)2(s) = Cr^2+(aq) + 2 OH^-(aq); Ksp = 2 x 10^-16
......................x.................2x

a) Ksp = [Cr^2+][OH^-]^2 = (x)(2x)^2 = 4x^3
4x^3 = 2 x 10^-16
x^3 = 5 x 10^-17
x = 3.68 x 10^-6 M This is the molar solubility of Cr(OH)2 in water

b) pH = 11. [H^+] = 10^-pH = 10^-11 = 1 x 10^-11; Kw = [H^+][OH^-] = 1 x 10^-14
[OH^-] = Kw/[H^+] = (1 x 10^-14)/(1 x 10^-11) = 1 x 10^-3 M OH^-

Ksp = [Cr^2+][OH^-]^2 = 2 x 10^-16
In this case the solubility of the Cr(OH)2 is equal to the [Cr^2+]
[Cr^2+] = Ksp/[OH^-]^2 = (2 x 10^-16)/(1 x 10^-3)^2 = 2 x 10^-10 M
molar solubility = 2 x 10^-10 M

c) 0.1M Cr(NO3)2 will give us a [Cr^2+] = 0.1M
In this case the molar solubility will be related to the [OH^-] in solution that is calculated from the Ksp equation. Remember, [Cr^2+] = [OH^-]/2 in the solubility equation.

[Cr^2+][OH^-]^2 = 2 x 10^-16
(0.1)[OH^-]^2 = 2 x 10^-16
[OH^-]^2 = 2 x 10^-15
[OH^-] = 4.5 x 10^-8 M For each Cr^2+ in solution 2 OH^- ions are produced by dissolving a Cr(OH)2 formula unit. Therefore, [Cr^2+] = [OH^-]/2 = 2.3 x 10^-8 M
molar solubility = 2.3 x 10^-8 M

Molar solubility, calculate Ksp help?

The molar solubility of MgCO3 is 1.8 x 10^-4 mol/L. What is the Ksp for this compound? Ksp = [Mg] [CO]^3 Ksp = [1.8 x 10^-4] [3 x 1.8 x 10^-4]^3 Ksp = 2.8 x 10^-14 apparently thats not correct and the answer is actually 3.2 x 10^-8 What am I doing wrong?

You have the right idea, you just made an early mistake. First it helps if you write out the ionization equation, which is:

MgCO3 --> Mg2+ + (CO3)2-

From what you have, it looks as if you thought there were 3 moles of CO to each mole of Mg, but the ion is (CO3)2-, not (CO)3.

Once you know that, you should get the answer. Instead of your Ksp, the new Ksp you get should be:
Ksp= [Mg2+][CO32-]

From there, they give you the molar solubility, which by molar ratios is the same as the concentration of each ion, and you can solve.

molar solubility Ksp question?

1) Use the molar solubility 1.55*10^-5 M in pure water to calculate Ksp for Ag2SO3. 2) Use the molar solubility 2.22*10^-8 M in pure water to calculate Ksp for Pd(SCN)2.

(1)
Ksp = [Ag+]^2 [SO3^-2]
[Ag+] = 2x1.55x10^-5
[SO3-2] = 1.55x10^-5
Ksp = 9.61x10^-10 x 1.55x10^-5 = 1.49x10^-14

(2)
Ksp = [Pb+2][SCN^-1]^2
[Pb+2] = 2.22x10^-8
[SCN-] = 4.44x10^-8
Ksp = 4.38x10^-23

How to calculate Ksp from molar solubility?

A.)If the molar solubility of CaF2 at 35degC is 1.24x10^-2mol/L, what is Ksp at this temperature?? B.) It is found that 1.1x10^-2g of SrF2 dissolves per 100mL of aqueous solution at 25degC. Calculate solubility product constant. Answers: A.)Ksp = 7.63x10^-9 B.) Ksp= 2.7x10^-9 But how?! Help! Thanks!

CaF2 --> Ca 2+ + 2F-
0.0124M CaF2 --> 0.0124M Ca 2+ + 0.0248M F-
Ksp = [Ca2+][F-]^2
Ksp = (0.0124)(0.0248)^2 = 7.63x10^-6

0.011g / 125.6g/mole = 8.76x10^-5moles
8.76x10^-5moles / 0.1L = 8.76x10^-4M

8.76x10^-4M SrF2 --> 8.76x10^-5M Sr 2+ and 1.75x10^-4M F-
Ksp = [Sr2+][F-]^2
Ksp = 8.75x10^-4 x (1.75x10^-4)^2 = 2.68x10^-11

Use the molar solubility in pure water to calculate Ksp for these three compounds.?

A) Use the molar solubility 1.08 x 10^-5 M in pure water water to calculate Ksp for BaCrO4. B) Use the molar solubility 1.55 x 10^-5 M in pure water water to calculate Ksp for Ag2SO3 C) Use the molar solubility 2.22 x 10^-8 M in pure water water to calculate Ksp for Pd(SCN)2.

A) Ksp = [Ba++][CrO4--]
For every x moles of BaCrO4 that dissolves in a liter of water, the concentration of Ba++ is x Molar, and the concentration of CrO4-- is x Molar. Therefore,
Ksp = (x)(x) = x² = (1.08 x 10^-5 M)² = 1.17 x 10^-10 M².

B) Ksp = [Ag]²[SO3--]
For every x moles of Ag2SO3 that dissolves in a liter of water, the concentration of Ag+ is 2x Molar, and the concentration of SO3-- is x Molar. Therefore,
Ksp = (2x)²(x) = 4x³ = (4)(1.55 x 10^-5 M)³ = 1.49 x 10^-14 M³.

C) Ksp = [Pd++][SCN-]²
In the same manner as question B),
Ksp = (x)(2x)² = 4x³ = (4)(2.22 x 10^-8 M)³ = 4.38 x 10^-23 M³.

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