150 , mL of 0.0008 , M ammonium sulphate is m | vedclass

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(A) First,calculate the final concentration of ions after mixing:Total volume = $150 \, mL + 50 \, mL = 200 \, mL$.For $[Ca^{2+}]$: $M_2 = \frac{M_1 V

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$< K_{SP}$

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(A) First,calculate the final concentration of ions after mixing:Total volume = $150 \, mL + 50 \, mL = 200 \, mL$.For $[Ca^{2+}]$: $M_2 = \frac{M_1 V_1}{V_2} = \frac{0.04 	imes 50}{200} = 0.01 \, M$.For $[SO_4^{2-}]$: $M_2 = \frac{M_1 V_1}{V_2} = \frac{0.0008 	imes 150}{200} = 0.0006 \, M$.Ionic product $(Q) = [Ca^{2+}] [SO_4^{2-}] = (0.01) 	imes (0.0006) = 6 	imes 10^{-6}$.Comparing with $K_{SP} = 2.4 	imes 10^{-5}$,we find $6 	imes 10^{-6} Therefore,$Q < K_{SP}$.

$150 \, mL$ 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 ($K_{SP} = 2.4 \times 10^{-5}$ for $CaSO_4$)

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