# Question a4cd9

Jun 3, 2017

No ppt forms ... ${K}_{\text{sp}} \left(= 2.4 x {10}^{-} 5\right)$ > ${Q}_{\text{sp}} \left(1.6 x {10}^{-} 5\right)$

#### Explanation:

$20 m l \left(0.01 M C a C {l}_{2}\right) + 20 m l \left(0.008 M N {a}_{2} S {O}_{4}\right)$
=> $\left(0.02\right) \left(0.01\right) \text{mole} C a C {l}_{2}$ + $\left(0.02\right) \left(0.008\right) \text{mole} N {a}_{2} S {O}_{4}$
=> $0.0002 \text{mole} C a C {l}_{2}$ + $0.00016 \text{mole} N {a}_{2} S {O}_{4}$

=> $N {a}_{2} S {O}_{4}$ is the limiting reagent => 1.6x10^-4" mole $C a S {O}_{4}$

$\left[C a S {O}_{4}\right]$ = 1.6x10^-4"# $\text{mole} C a S {O}_{4}$ / $\text{0.040L Solution}$ = $0.004 M$

0.004M$C a S {O}_{4} r i g h t \le f t h a r p \infty n s$ +0.004M$\left(C {a}^{+ 2}\right)$ + 0.004M$\left(S {O}_{4}^{2 -}\right)$

${Q}_{s p} = \left[C {a}^{+ 3}\right] \left[S {O}_{4}^{2 -}\right] = \left(0.004 M\right) \left(0.004 M\right)$ = $1.6 x {10}^{-} 5$

${K}_{s p} = 2.4 x {10}^{-} 5 > {Q}_{s p} = 1.6 x {10}^{-} 5$ => No Ppt will form.