Name: Basic Buffers
Uploaded: 2019-07-19
Duration: 6 min 4 s
Description: This video explains buffer action as the resistance to change in pH of a solution on the addition of an acid or a base. Learn of the types of buffer solution...

Question

What is the relationship between molar solubility and solubility product for salts given below
i.

{Ag}_{2} {CrO}_{4}

ii.

{Ca}_{3} {({PO}_{4})}_{2}

iii.

Cr (OH)_{3}

.

Accepted Answer

The solubility of ionic compounds that dissociate and form cations and anions in water varies to a great extent. Some compounds are highly soluble and can even absorb moisture from the atmosphere whereas, on the other hand, the others are highly insoluble.

${Ag}_{2} {CrO}_{4} ⇌ 2 {Ag}^{+} + {CrO}_{4}^{2 -} . K_{sp} = {[{Ag}^{+}]}^{2} \times [{CrO}_{4}^{2 -}] . K_{sp} = {(2 S)}^{2} \times (S) = 4 S^{3} . S = {(\frac{K_{sp}}{4})}^{\frac{1}{3}} mol {dm}^{- 3} .$

${Ca}_{3} {({PO}_{4})}_{2} ⇌ 3 {Ca}^{2 +} + 2 {PO}_{4}^{3 -} . K_{sp} = {[{Ca}^{2 +}]}^{3} \times {[{PO}_{4}^{3 -}]}^{2} . K_{sp} = {(3 S)}^{3} \times {(2 S)}^{2} = 108 S^{5} . S = {(\frac{K_{sp}}{108})}^{\frac{1}{5}} mol {dm}^{- 3} .$

$Cr {(OH)}_{3} ⇌ {Cr}^{3 +} + 3 {OH}^{-} . K_{sp} = [{Cr}^{3 +}] \times {[{OH}^{-}]}^{3} . K_{sp} = (S) \times {(3 S)}^{3} = 27 S^{4} . S = {(\frac{K_{sp}}{27})}^{\frac{1}{4}} mol {dm}^{- 3} .$

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