Question

The solubility product constants of $Ag_{2}CrO_{4}$ and AgBr are 1.1 × $10^{-12}$ and 5.0 × $10^{-13}$ respectively. Calculate the ratio of the molarities of their saturated solutions.

Answer 1

"For $A{ g }_{ 2 }C{ rO }_{ 4 },$

At equilibrium the solubility of the reaction is as follows.

$A{ g }_{ 2 }Cr{ O }_{ 4 }\quad \rightleftharpoons \quad 2A{ g }^{ + }(aq)\quad +\quad Cr{ O }_{ 4 }^{ 2- }(aq)$

Let the solubility of $A{ g }_{ 2 }Cr{ O }_{ 4 } be S mol\{ L }^{ -1 }$

From the reaction

$[{ Ag }^{ + }]\quad =\quad 2S;\quad [Cr{ O }_{ 4 }^{ 2- }]\quad =\quad S$

${ K }_{ sp\quad}=\quad { [Ag }^{ + }][{ CrO }_{ 4 }^{ - }]$

$=\quad { (2S) }^{ 2 }(S)$

${ K }_{ sp\quad}=\quad { 4S }^{ 3 }$

Or

${ 4S }^{ 3 }\quad =\quad { K }_{ sp\quad}$

Or

${ S }^{ 3 }\quad =\quad \frac { { K }_{ SP } }{ 4 }$

Or

$S\quad =\quad { \left[ \frac { { K }_{ SP } }{ 4 } \right]}^{ \sfrac { 1 }{ 3 }}$

$\therefore \quad S\quad =\quad { \left( \frac { 1.1\quad \times \quad { 10 }^{-12}}{4}\right)}^{ 1/3 }\quad =\quad 0.65\quad \times \quad { 10 }^{ -4 }\quad mol{ L }^{ -1 }$

Thus,${ M }_{ { Ag }_{ 2 }{ CrO }_{ 4 } }\quad =\quad 0.65\quad \times { \quad 10 }^{ -4 }\quad mol.{ L }^{ -1 }$

For AgBr, the solubility equilibrium is

$AgBr\quad \rightleftharpoons \quad { Ag }^{ + }(aq)\quad +\quad { Br }^{ - }(aq)$

Let the solubility of AgBr be S mol.${ L }^{ - }$

$\therefore \quad { [Ag }^{ + }]\quad =\quad S;\quad { [Br }^{ - }]\quad =\quad S$

${ K }_{ SP }\quad =\quad { [Ag }^{ + }]{ [Br }^{ - }]$

$=\quad (S)(S)$

${ K }_{ SP }\quad =\quad { S }^{ 2 }$

Or

${ S }^{ 2 }\quad =\quad { K }_{ SP }$

Or

$S\quad =\quad { (K }_{ SP }{ ) }^{ 1/2 }$

Or

$S\quad =\quad (5.0\quad \times \quad 10^{ -13 }{ ) }^{ 1/2 }$

$S\quad =\quad 0.707\quad \times \quad { 10 }^{ -6 }\quad mol.{ L }^{ -1 }$

The ratio of molarities of their saturated solutions $=\quad \frac { { M }_{ { Ag }_{ 2 }{ CrO }_{ 4 } } }{ { M }_{ { Ag }Br } } \quad =\quad \frac { 0.65\quad \times \quad { 10 }^{ -4 } }{ 0.707\quad \times \quad { 10 }^{ -6 } } \quad =\quad 91.9$

Since the solubility of $A{ g }_{ 2 }Cr{ O }_{ 4 }$ is more than that of AgBr, so the former is more soluble."

Answer 2

"For A{ g }_{ 2 }C{ rO }_{ 4 },Ag

2

CrO

4

,

At equilibrium the solubility of the reaction is as follows.

A{ g }_{ 2 }Cr{ O }_{ 4 }\quad \rightleftharpoons \quad 2A{ g }^{ + }(aq)\quad +\quad Cr{ O }_{ 4 }^{ 2- }(aq)Ag

2

CrO

4

⇌2Ag

+

(aq)+CrO

4

2−

(aq)

Let the solubility of

From the reaction

[{ Ag }^{ + }]\quad =\quad 2S;\quad [Cr{ O }_{ 4 }^{ 2- }]\quad =\quad S[Ag

+

]=2S;[CrO

4

2−

]=S

{ K }_{ sp\quad}=\quad { [Ag }^{ + }][{ CrO }_{ 4 }^{ - }]K

sp

=[Ag

+

][CrO

4

−

]

=\quad { (2S) }^{ 2 }(S)=(2S)

2

(S)

{ K }_{ sp\quad}=\quad { 4S }^{ 3 }K

sp

=4S

3

Or

{ 4S }^{ 3 }\quad =\quad { K }_{ sp\quad}4S

3

=K

sp

Or

{ S }^{ 3 }\quad =\quad \frac { { K }_{ SP } }{ 4 }S

3

=

4

K

SP

Or

\therefore \quad S\quad =\quad { \left( \frac { 1.1\quad \times \quad { 10 }^{-12}}{4}\right)}^{ 1/3 }\quad =\quad 0.65\quad \times \quad { 10 }^{ -4 }\quad mol{ L }^{ -1 }∴S=(

4

1.1×10

−12

)

1/3

=0.65×10

−4

molL

−1

Thus,{ M }_{ { Ag }_{ 2 }{ CrO }_{ 4 } }\quad =\quad 0.65\quad \times { \quad 10 }^{ -4 }\quad mol.{ L }^{ -1 }M

Ag

2

CrO

4

=0.65×10

−4

mol.L

−1

For AgBr, the solubility equilibrium is

AgBr\quad \rightleftharpoons \quad { Ag }^{ + }(aq)\quad +\quad { Br }^{ - }(aq)AgBr⇌Ag

+

(aq)+Br

−

(aq)

Let the solubility of AgBr be S mol.{ L }^{ - }L

−

\therefore \quad { [Ag }^{ + }]\quad =\quad S;\quad { [Br }^{ - }]\quad =\quad S∴[Ag

+

]=S;[Br

−

]=S

{ K }_{ SP }\quad =\quad { [Ag }^{ + }]{ [Br }^{ - }]K

SP

=[Ag

+

][Br

−

]

=\quad (S)(S)=(S)(S)

{ K }_{ SP }\quad =\quad { S }^{ 2 }K

SP

=S

2

Or

{ S }^{ 2 }\quad =\quad { K }_{ SP }S

2

=K

SP

Or

S\quad =\quad { (K }_{ SP }{ ) }^{ 1/2 }S=(K

SP

)

1/2

Or

S\quad =\quad (5.0\quad \times \quad 10^{ -13 }{ ) }^{ 1/2 }S=(5.0×10

−13

)

1/2

S\quad =\quad 0.707\quad \times \quad { 10 }^{ -6 }\quad mol.{ L }^{ -1 }S=0.707×10

−6

mol.L

−1

The ratio of molarities of their saturated solutions =\quad \frac { { M }_{ { Ag }_{ 2 }{ CrO }_{ 4 } } }{ { M }_{ { Ag }Br } } \quad =\quad \frac { 0.65\quad \times \quad { 10 }^{ -4 } }{ 0.707\quad \times \quad { 10 }^{ -6 } } \quad =\quad 91.9=

M

AgBr

M

Ag

2

CrO

4

=

0.707×10

−6

0.65×10

−4

=91.9

Since the solubility of A{ g }_{ 2 }Cr{ O }_{ 4 }Ag

2

CrO

4

is more than that of AgBr, so the former is more soluble."