State solubility product principle. A sample of AgCl is treated with 5mL of 2M Na2CO3 to produce Ag2CO3. The remaining solution contained 0.03g of Cl- pre liter. Calculate the solubility product of AgCl.

Question: 

(a) State solubility product principle. A sample of AgCl is treated with 5mL of 2M Na₂CO₃ to produce Ag₂CO₃. The remaining solution contained 0.03g of Cl^- pre liter. Calculate the solubility product of AgCl.

(b)What volume of 36% HCl having specific gravity 1.18 is required to prepare 1liter of its solution having p^H 1?

Solution:

Solubility Product Principle:

The solubility product principle states that for a sparingly soluble salt in water, the product of the concentrations of its constituent ions, raised to the power of their stoichiometric coefficients in the balanced chemical equation, is a constant at a given temperature. This constant is called the solubility product (K_sp) and is a measure of the extent to which the salt dissolves in water.

Given:

[Cl^-]=[NaCl]= $\frac{{0.03}}{{35.5}}$= 8.4 × 10^-4

[CO₃^-2]=[ Na₂CO₃]=2M

Reaction:

Na₂CO₃ + 2AgCl → 2NaCl + Ag₂CO₃

Reaction	Na2CO3	2AgCl	→	2NaCl	Ag2CO3
Initial	2M			0
At Equilibrium	(2-8.4 × 10-4)			8.4 × 10-4

[Ag]⁺²[CO₃^-2]=$ {K_{s{p_{(A{g_2}C{O_3})}}}}$

Therefore, \[\;[A{g^{ + 2}}] = \sqrt {\frac{{{K_{s{p_{(A{g_2}C{O_3})}}}}}}{{\left[ {C{O_3}^{ - 2}} \right]}}}  = \sqrt {\frac{{8.2 \times {{10}^{ - 12}}}}{2}}  = 2.02 \times {10^{ - 6}}\]

And, ${K_{s{p_{(AgCl)}}}} = [A{g^ + }][C{l^ - }] = 2.02 \times {10^{ - 6}} \times 8.4 \times {10^{ - 4}} = 1.7 \times {10^{ - 9}}$

(b)

Concentration of H⁺ [H⁺]= 0.1

Moles of [H⁺]= Molarity x Volume = 0.1 x 1 = 0.1

HCl→H⁺ + Cl^-

Moles of H⁺ = Moles of HCl= 0.1=10^-1

Mass of HCl = Mole x Molecular Weight = 0.1 x 36.5 = 3.65gram

This mass is equal to 36% of HCl.

Mass of Solution = $\frac{{3.65}}{{36}} \times 100$= 10.13gram

Volume = $\frac{{Mass}}{{Density}}$

$\frac{{10.13}}{{1.18}}$= 8.59 liter