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

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DOLU






SOLUBILITY OF.

$Ag_{2}CrO_{4}$ is more soluble than $AgBr$