Question

A certain town gets its water from an underground aquifer that contains water in equilibrium with limestone (calcium carbonate). One resident leaves town of vacation and in the process, leave a cup of water (200mL) on the kitchen table. When he returns, the water has evaporated, leaving only calcium carbonate in the cup.(Hint: you need to the Ksp from the appendix!) a. What is the molar solubility of calcium carbonate in the water? b. How much calcium carbonate in remaining in the cup? c. How much calcium is present in grams. d. For insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility. If the residents wants to clean the cup, is he better doing so with an acidic cleaner (like lemon juice) or a basic solution (like soap or baking soda)? (Consider the common ion effect

* Calcium carbonate Ksp is 3.33x10^-9 *

Answer 1

since there are no units for K_sp iam assuming it is M² or (m/L)²

a)Calcium Carbonate dissolves in water according to below equation

CaCO₃$\rightarrow$ Ca²⁺ + CO₃^2-

Let the solubility of CaCO₃ be s M

then concentration of Ca²⁺ and  CO₃^2- in the solution will be s M (since they are formed in equal concentration with CaCO₃)

therefore the solubility product = K_sp = [Ca²⁺][ CO₃^2-]

= (s)$\times$(s)

= s²

given solubility product = K_sp = 3.33 $\times$ 10^-9

therefore s² = 3.33 $\times$ 10^-9

s =

V3.33 × 10

s = 5.77 $\times$ 10^-8 M

therefore molar solubility of calcium carbonate in water = 5.77 $\times$ 10^-8 M

b)

Volume of water in the cup = 200mL = 0.2 L

number of moles of calcium carbonate in cup = (molar solubility of calcium carbonate )$\times$(Volume of water in the cup)

= 5.77 $\times$ 10^-8 M $\times$ 0.2 L

= 1.154 $\times$ 10^-8 moles

mass of calcium carbonate remaining in the cup = (number of moles of calcium carbonate in cup)$\times$(molar mass of calcium carbonate)

= (1.154 $\times$ 10^-8 )$\times$(100.08 g)

= 1.154 $\times$ 10^-6g

c)

one mole of CaCO₃ contains one mole of calcium atoms

therefore number of moles of calcium present in the cup = number of moles of calcium carbonate in cup

= 1.154 $\times$ 10^-8 moles

mass of calcium remaining in the cup = (number of moles of calcium in the cup)$\times$(molar mass of calcium) = (1.154 $\times$ 10^-8 )$\times$(40.078 g)

= 4.625 $\times$ 10^-7 g

d)

Acidic cleaners have high concentrations of H⁺ ions

these H⁺ ions react with calcium carbonate to form calcium bicarbonate which is more soluble in water as a result can be washed away easily.

CaCO₃  $\rightleftharpoons$ Ca²⁺ + CO₃^2- + H⁺  $\rightleftharpoons$ HCO₃^-+ Ca²⁺