The solubility product of Ca SO 4 is 2-4- 10 -5- When 100mL of 0-01M CaCl2 and 100mL of 0-02M Na2SO4 are mixed then-

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Common Ion Effect

The solubilit...

Question

The solubility product of $C a {S O}_{4}$ is $2.4 \times 10^{- 5}$ . When $100 m L$ of $0.01 M$ $C a {C l}_{2}$ and $100 m L$ of $0.02 M$ ${N a}_{2} {S O}_{4}$ are mixed then:

{N a}_{2} {S O}_{4}

will precipitate

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Both will precipitate

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C a {S O}_{4}

will precipitate

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None will precipitate

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Solution

The correct option is B $C a {S O}_{4}$ will precipitate
Molarity of $C a^{2 +}$ in the mixture.

$[C a^{2 +}] = 0.01 \times 100 / 200 = 0.005 M$

Molarity of $S O_{4}^{2 -}$

$[S O_{4}^{2 -}] = 0.02 \times 100 / 200 = 0.01 M$

$Q_{i o n i c p r o d u c t} = [C a^{2 +}] [S O_{4}^{2 -}] = 0.005 \times 0.01 = 5 \times 10^{- 5}$

As $Q_{i p} > K_{s p}$ so $C a S O_{4}$ will precipitate.

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A solution prepared by mixing $200 m l$ of $0.025 M$ $C a C l_{2}$ and $400 m l$ of $0.15 M$ $N a_{2} S O_{4}$ ? Given $K_{s p}$ of $C a S O_{4}$ = $2.4 \times 10^{- 5}$ Then

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The solubility product of CaSO4 is 2.4×10−5. When 100mL of 0.01M CaCl2 and 100mL of 0.02M Na2SO4 are mixed then:

The correct option is B CaSO4 will precipitateMolarity of Ca2+ in the mixture.[Ca2+]=0.01×100/200=0.005MMolarity of SO2−4[SO2−4]=0.02×100/200=0.01MQionicproduct=[Ca2+][SO2−4]=0.005×0.01=5×10−5As Qip>Ksp so CaSO4 will precipitate.

The solubility product of $C a {S O}_{4}$ is $2.4 \times 10^{- 5}$ . When $100 m L$ of $0.01 M$ $C a {C l}_{2}$ and $100 m L$ of $0.02 M$ ${N a}_{2} {S O}_{4}$ are mixed then: