Question

A student dissolves calcium carbonate solids, CaCO_3(s) in distilled water. K_sp = 5.0 times 10^-9 a) Calculate the molar solubility of calcium carbonate in distilled water. Express your answer in gram per litre b) How would the answers in a) change if the student dissolved CaCO_3 in 0.01 mol/L solution of Na_2CO_3(aq) c) The student mixes 0.0140 g of CaO solids with 0.0265 g of Na_2CO_3 solids in 1.0 L of water solids are soluble in water). Predict whether precipitate will form.

Answer 1

A) CaCO3 (s) <---> Ca²⁺ (aq) + CO₃^2-(aq)

Now, the molar ratio of ions is 1:1, so we can say that the concentrations of both the ions is equal.

Now, Ksp = [Ca2+][CO₃^2-].

Ksp = x . x

Ksp = X²

(Ksp)^1/2 = X

X = (5.0 X 10^-9)^1/2 = 0.0000707106 = 7.07 x 10^-5 moles/L

converting moles to g

moles x molecular weight = given weight

7.07x 10-5 mol X 100.0869 g/mol = 707.61x10^-5 = 7.07 x 10^-3 g/l

B) answer to this would take care of the common ion effect on molar solubility.

0.01mol/L of sodium carbonate

now,

[Ca²⁺] = x

[CO₃^2-] = 0.01+x

now, solubility product equation is

Ksp = (x )(0.01+x)

5.0X 10^-9 = 0.01x + X²

0.01x + X² - 5.0X 10^-9 =0

X² +0.01x - 5.0X 10^-9 =0

Solving quadratic

x = -0.01+/-[(0.01)² - 4(1)(- 5.0X 10^-9)]^1/2/ 2

x = 4.99 x 10^-7 mol/L

so, now the solubility is much lesser and hence the solid would precipitate out more quickly on addition of a common as seen here

C) CaO(s) + H₂O(l) + Na₂CO₃(s) ---> CaCO₃(s) + 2 NaOH(aq)

moles of CaO =0.0140/56 = 0.00025 moles

moles of Na2CO3 =0.0265/106 =0.00025 moles

moles of CaCo3 formed would be = 0.00025 moles

if ion product < Ksp no precipitation occurs

ion product is 6.25 x 10^-8 and Ksp = 5.00 X 10^-9

so, here precipitation will take place.