150, mL of 0.0008, M ammonium sulphate is mixed with 50, mL of 0.04, M calcium nitrate. ionic produc

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6-2.Equilibrium-II (Ionic Equilibrium)

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$150\, mL$ of $0.0008\, M$ ammonium sulphate is mixed with $50\, mL$ of $0.04\, M$ calcium nitrate. The ionic product of $CaSO_4$ will be ($K_{SP}= 2.4\times10^{-5}$ for $CaSO_4$)

A

$ < K_{SP}$

B

$ > K_{SP}$

C

$ \approx K_{SP}$

D

None of these

Solution

For $\left[\mathrm{Ca}^{+2}\right], \mathrm{M}_{1} \mathrm{V}_{1}=\mathrm{M}_{2} \mathrm{V}_{2}$

$0.04 \times 50=M_{2} \times 200$

${M}_{2}=0.01$

Similarly, for $\left[\mathrm{SO}_{4}^{-2}\right]$

$0.0008 \times 150=\mathrm{M}_{2} \times 200$

$\mathrm{M}_{2}=0.0006$

ionic product $(\mathrm{Q})=\left[\mathrm{Ca}^{+2}\right] \times\left[\mathrm{SO}_{4}^{-2}\right]$

$=6 \times 10^{-6}$

So, $\quad \quad {Q} < {K}_{\mathrm{SP}}$

Standard 11

Chemistry

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