Solveeit Logo
Login

Question

Question: The solubility of silver sulphate in the water at \(100{\,^{\text{o}}}{\text{C}}\) is approximately ...

The solubility of silver sulphate in the water at 100 oC100{\,^{\text{o}}}{\text{C}} is approximately 1.41.4g per 100100 mL. What is the solubility product of this salt at 100 oC100{\,^{\text{o}}}{\text{C}}?
A. 5.7×10−85.7 \times {10^{ - 8}}
B. 3.5×10−73.5 \times {10^{ - 7}}
C. 8.3×10−68.3 \times {10^{ - 6}}
D. 4.1×10−54.1 \times {10^{ - 5}}
E. 3.6×10−43.6 \times {10^{ - 4}}

Explanation

Solution

To solve this we must know that the irregularities formed in the arrangement of constituent particles are known as defects. Frenkel defect occurs when there is no disturbance in the stoichiometry of the solid. Frenkel defect occurs in ionic solids in which there is a large difference in the size of the ions.

Complete step-by-step answer:
We will determine the mole of silver sulphate as follows:
mole  =  mass molar mass{\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{\,{\text{molar}}\,{\text{mass}}}}
Molar mass of silver sulphate is 132132 g/mol.
On substituting 1.41.4g for mass and 132132 g/mol for molar mass,
mole  =  1.4 g 312 g/mol{\text{mole}}\,{\text{ = }}\,\dfrac{{1.4\,{\text{g}}}}{{\,312\,{\text{g/mol}}}}
mole  =  4.5 ×10−3 mol{\text{mole}}\,{\text{ = }}\,4.5\, \times {10^{ - 3}}\,{\text{mol}}
So, the mole of silver sulphate is 4.5 ×10−34.5\, \times {10^{ - 3}} mole.
We will convert the volume of solution from mL to L as follows:
1000 mL  =  1 L1000\,{\text{mL}}\,{\text{ = }}\,{\text{1 L}}
100 mL=0.1 L100\,{\text{mL}} = 0.1\,{\text{L}}
We will determine the molarity as follows:
molarity   =  moles of solutevolume of solution{\text{molarity}}\,\,{\text{ = }}\,\dfrac{{{\text{moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{volume}}\,{\text{of}}\,{\text{solution}}}}

On substituting 0.10.1L for volume of solution and 4.5 ×10−34.5\, \times {10^{ - 3}}mol for mole of silver sulphate.
Molarity  =  4.5 ×10−3 mol0.1  L{\text{Molarity}}\,{\text{ = }}\,\dfrac{{4.5\, \times {{10}^{ - 3}}\,{\text{mol}}}}{{0.1\,\,{\text{L}}}}
Molarity  =  0.045 M{\text{Molarity}}\,{\text{ = }}\,0.045\,{\text{M}}
So, the molarity (concentration) or solubility of silver sulphate solution is 0.0450.045M.

The solubility product constant represents the product of concentrations of constituting ions of an ionic compound at equilibrium each raised number of ions as power.
An ionic compound dissociates into the water as follows:
AxBy ⇄  xA + Y  +  yB−x{{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\, \rightleftarrows \,\,{\text{x}}{{\text{A}}^{{\text{ + Y}}}}\,{\text{ + }}\,{\text{y}}{{\text{B}}^{ - x}}
Where,
AB{\text{AB}} is an ionic compound.
The general representation for the solubility product of an ionic compound is as follows:
Ksp  =  (xA + y )x ×  (yA−x )y{{\text{K}}_{{\text{sp}}}}\,{\text{ = }}\,{\left( {{\text{x}}{{\text{A}}^{{\text{ + y}}}}\,} \right)^{{x}}}\, \times \,\,{\left( {{\text{y}}{{\text{A}}^{ - x}}\,} \right)^{\text{y}}}
Ksp  =  xSx × ySy{{\text{K}}_{{\text{sp}}}}\,{\text{ = }}\,{\text{x}}{{\text{S}}^{\text{x}}}\,{{ \times }}\,{\text{y}}{{\text{S}}^{\text{y}}}
Where,
Ksp{{\text{K}}_{{\text{sp}}}}is the solubility product constant.
S{\text{S}}is the solubility of each ion.
xx and yy represents the number of ions.
The compound silver sulfide dissociates into the water as follows:
Ag2SO4⇄  2 Ag +  +   SO42−{\text{A}}{{\text{g}}_2}{\text{S}}{{\text{O}}_4} \rightleftarrows \,\,2\,{\text{A}}{{\text{g}}^{\text{ + }}}{\text{ + }}\,\,{\text{SO}}_4^{2 - }
Silver sulphate in water produces two silver ions and one sulphate ion.
The solubility product for silver sulphate is represented as follows:
Ksp =  [2Ag + ]2[Br−]{{\text{K}}_{{\text{sp}}}}\, = \,\,{\left[ {{\text{2A}}{{\text{g}}^{\text{ + }}}} \right]^2}\left[ {{\text{B}}{{\text{r}}^ - }} \right]
Ksp =  [2S]2[S]{{\text{K}}_{{\text{sp}}}}\, = \,\,{\left[ {{\text{2S}}} \right]^2}\left[ {\text{S}} \right]
Ksp =  4S3{{\text{K}}_{{\text{sp}}}}\, = \,\,{\text{4}}{{\text{S}}^3}
On substituting 0.0450.045 M for solubility in the above equation,
Ksp =  4×(0.045)3{{\text{K}}_{{\text{sp}}}}\, = \,\,{\text{4}} \times {\left( {0.045} \right)^3}
Ksp =  3.6×10−4{{\text{K}}_{{\text{sp}}}}\, = \,\,3.6 \times {10^{ - 4}}
Therefore, option (E) Ksp =  3.6×10−4{{\text{K}}_{{\text{sp}}}}\, = \,\,3.6 \times {10^{ - 4}}, is correct.

Note: By the addition of atomic mass the molar mass is determined. Solubility tells the concentration of solid compounds dissolved in water. Concentration in the determination of solubility products is taken in terms of molarity. The formula of solubility product constant depends upon the number of ions produced by the ionic compounds in the water. Molar solubility represents the number of ions dissolved per liter of solution. Here, solubility represents the number of ions dissolved in a given amount of solvent.