1.What is the molar solubility of CaCO3 (Ksp = 2.5 x 10 -9 ) in an aqueous solut

Question

1.What is the molar solubility of CaCO3 (Ksp = 2.5 x 10-9) in an aqueous solution where the carbonate ion concentration is 0.30 M at 25°C?

2. Two solutions are mixed; one solution contains Al3+ and the other solution contains OH-. If, at the instant of mixing, but before any precipitate forms, [Al3+] = 3.2 x 10-10 M and [OH-] = 5.6 x 10-8 M, which of the following statements is true? The Ksp value for Al(OH)3 is 4.6 x 10-33 at 25°C?

Which salt is the least soluble in water at 25°C?

No precipitate forms because Qsp > Ksp A precipitate forms because Qsp > Ksp No precipitate forms because Qsp < Ksp A precipitate forms because Qsp < Ksp


3.Which salt is the least soluble in water at 25°C?

Which salt is the least soluble in water at 25°C?

Mg(OH)2 (Ksp = 1.8 x 10-11) PbI2 (Ksp = 6.5 x 10-9) CaF2 (Ksp = 3.4 x 10-11) CaC2O4 (Ksp = 2.3 x 10-9)

Explanation / Answer

Answer – 1)

We are given, [CO32-] = 0.30 M, Ksp of CaCO3 = 2.5*10-9

We know the formula for calculating the molar solubility

Ksp = [Ca2+] [CO32-]

2.5*10-9 = x * 0.30 M

x = 2.5*10-9 / 0.30 M

   = 9.67*10-9 M

So, molar solubility of CaCO3 is 9.67*10-9 M

2) Given, [Al3+] = 3.2 x 10-10 M , [OH-] = 5.6 x 10-8 M

Ksp value for Al(OH)3 = 4.6 x 10-33

We need to calculate the Qsp for the Al(OH)3

We know the Qsp is the product of the initial concentration

So, Qsp expression is

Qsp = [Al3+] [OH-]3

=   (3.2 x 10-10 M) (5.6 x 10-8 M)3

= 5.62*10-32

So we the Qsp > Ksp , so there is precipitate will be formed

So answer is A precipitate forms because Qsp > Ksp

3) We know as the Ksp means solubility product increase the molar solubility increase, so we need to look which one has the highest Ksp value and that will be dissolve most and the lowest one is dissolved least.

The lowest Ksp for the Mg(OH)2 and it is Ksp = 1.8 x 10-11, so this one is the least soluble in the water. So answer is Mg(OH)2 (Ksp = 1.8 x 10-11)

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