Question
Write expression for solubility and solubility product of following sparingly soluble salts :
$(1) \ AgBr\ (2) \ PbI _2\ (3) \ Al ( OH )_3$
$(1) \ AgBr\ (2) \ PbI _2\ (3) \ Al ( OH )_3$
✓
Answer
In general, for a sparingly soluble salt $A_x B_y$,
$K _{ sp }= x ^{ x } \cdot y ^{ y } \cdot( S )^{ x + y }$
$(1)$ For binary electrolyte $(BA$ type$)$ :
$ AgBr _{( s )} {\rightleftharpoons} \underset{x=1}{ Ag _{( aq )}^{+}}+\underset{y=1}{ Br _{( aq )}^{-}}$
$ \therefore K_{ sp } =x^x \cdot y^y \cdot(S)^{x+y}$
$ =1^1 \times 1^1 \times( S )^{1+1}$
$ =S^2$
$\therefore S=\sqrt{K_{\text {sp }}} mol dm ^{-3}$
$\text { (2) } PbI _{2(s)} \rightleftharpoons \underset{x=1}{ Pb _{\text {(aq) }}^{2+}}+ \underset{y=2}{2 I _{( aq )}^{-}}$
$\therefore K_{ sp }=x^x \cdot y^y \cdot(S)^{x+y}$
$=1^1 \times 2^2 \times( S )^{1+2}$
$=4 S^3$
$\therefore S=\left(\frac{K_{ sp }}{4}\right)^{\frac{1}{3}} \ mol \ dm ^{-3}$
$(3) Al ( OH )_{3(s)} \rightleftharpoons Al _{( aq )}^{3+}+3 OH _{( aq )}^{-}$
$K_{ sp }= {\left[ Al ^{3+}\right] \times\left[ OH ^{-}\right]^3 }$
$\therefore K_{ sp } =x^x \cdot y^y .(S)^{x+y}$
$ =1^1 \times 3^3 \times(S)^{1+3}$
$ =27 S^4$
$ \therefore S=\left(\frac{K_{ sp }}{27}\right)^{\frac{1}{4}}\ mol \ dm ^3$
$K _{ sp }= x ^{ x } \cdot y ^{ y } \cdot( S )^{ x + y }$
$(1)$ For binary electrolyte $(BA$ type$)$ :
$ AgBr _{( s )} {\rightleftharpoons} \underset{x=1}{ Ag _{( aq )}^{+}}+\underset{y=1}{ Br _{( aq )}^{-}}$
$ \therefore K_{ sp } =x^x \cdot y^y \cdot(S)^{x+y}$
$ =1^1 \times 1^1 \times( S )^{1+1}$
$ =S^2$
$\therefore S=\sqrt{K_{\text {sp }}} mol dm ^{-3}$
$\text { (2) } PbI _{2(s)} \rightleftharpoons \underset{x=1}{ Pb _{\text {(aq) }}^{2+}}+ \underset{y=2}{2 I _{( aq )}^{-}}$
$\therefore K_{ sp }=x^x \cdot y^y \cdot(S)^{x+y}$
$=1^1 \times 2^2 \times( S )^{1+2}$
$=4 S^3$
$\therefore S=\left(\frac{K_{ sp }}{4}\right)^{\frac{1}{3}} \ mol \ dm ^{-3}$
$(3) Al ( OH )_{3(s)} \rightleftharpoons Al _{( aq )}^{3+}+3 OH _{( aq )}^{-}$
$K_{ sp }= {\left[ Al ^{3+}\right] \times\left[ OH ^{-}\right]^3 }$
$\therefore K_{ sp } =x^x \cdot y^y .(S)^{x+y}$
$ =1^1 \times 3^3 \times(S)^{1+3}$
$ =27 S^4$
$ \therefore S=\left(\frac{K_{ sp }}{27}\right)^{\frac{1}{4}}\ mol \ dm ^3$
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