Question

How many grams of CaCO3 will dissolve in 3.00x10^2 mL of 0.0580 M Ca(NO3)2?

The Ksp for CaCO3 is 8.70x10-9

Answer 1

Ca(NO3)2 here is Strong electrolyte

It will dissociate completely to give [Ca2+] = 0.058 M

At equilibrium:

CaCO3 <----> Ca2+ + CO32-

   5.8*10^-2 +s s

Ksp = [Ca2+][CO32-]

8.7*10^-9=(5.8*10^-2 + s)*(s)

Since Ksp is small, s can be ignored as compared to 5.8*10^-2

Above expression thus becomes:

8.7*10^-9=(5.8*10^-2)*(s)

8.7*10^-9= 5.8*10^-2 * 1(s)^1

s = 1.5*10^-7 M

Molar mass of CaCO3,

MM = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

Molar mass of CaCO3= 100.09 g/mol

s = 1.5*10^-7 mol/L

To covert it to g/L, multiply it by molar mass

s = 1.5*10^-7 mol/L * 100.09 g/mol

s = 1.501*10^-5 g/L

volume = 3.00*10^2 mL = 0.300 mL

Now use:

mass soluble = s* volume

= (1.501*10^-5 g/L) * 0.300 L

= 4.50*10^-6 g

Answer: 4.50*10^-6 g