When 150 mL of 0.0008 M ammonium sulfate so | vedclass

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(A) First,calculate the final concentrations of the ions after mixing.Total volume $= 150 \ mL + 50 \ mL = 200 \ mL$.For $Ca^{2+}$ ions from $Ca(NO_3)

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$< K_{sp}$

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(A) First,calculate the final concentrations of the ions after mixing.Total volume $= 150 \ mL + 50 \ mL = 200 \ mL$.For $Ca^{2+}$ ions from $Ca(NO_3)_2$:$M_2 = \frac{M_1 V_1}{V_2} = \frac{0.04 \ M 	imes 50 \ mL}{200 \ mL} = 0.01 \ M$.For $SO_4^{2-}$ ions from $(NH_4)_2SO_4$:$M_2 = \frac{M_1 V_1}{V_2} = \frac{0.0008 \ M 	imes 150 \ mL}{200 \ mL} = 0.0006 \ M$.Ionic product $(Q) = [Ca^{2+}][SO_4^{2-}] = (0.01) 	imes (0.0006) = 6 	imes 10^{-6}$.Comparing $Q$ with $K_{sp}$:$Q = 6 	imes 10^{-6}$ and $K_{sp} = 2.4 	imes 10^{-5}$.Since $6 	imes 10^{-6} < 2.4 	imes 10^{-5}$,the ionic product is $< K_{sp}$.

When $150 \ mL$ of $0.0008 \ M$ ammonium sulfate solution is mixed with $50 \ mL$ of $0.04 \ M$ calcium nitrate solution,given $K_{sp} = 2.4 \times 10^{-5}$ for $CaSO_4$,what is the ionic product of $CaSO_4$?

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