Question

How many grams of CaCO3 will dissolve in 2.60 × 102 mL of 0.0560 M Ca(NO3)2? The Ksp for CaCO3 is 8.70 × 10−9.

Answer 1

Ca(NO3)2 here is Strong electrolyte

It will dissociate completely to give [Ca2+] = 0.056 M

At equilibrium:

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

   5.6*10^-2 +s s

Ksp = [Ca2+][CO32-]

8.7*10^-9=(5.6*10^-2 + s)*(s)

Since Ksp is small, s can be ignored as compared to 5.6*10^-2

Above expression thus becomes:

8.7*10^-9=(5.6*10^-2)*(s)

8.7*10^-9= 5.6*10^-2 * 1(s)^1

s = 1.554*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 = 1.554*10^-7 mol/L

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

s = 1.554*10^-7 mol/L * 100.09 g/mol

s = 1.555*10^-5 g/L

Now use:

mass of CaCO3 = s * volume

= 1.555*10^-5 g/L * 0.260 L

= 4.04*10^-6 g

Answer: 4.04*10^-6 g