Question #a4cd9

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Question #a4cd9

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Answer 1 · 2017-06-03T07:01:48

No ppt forms ... $K_{sp} (= 2.4 x 10^{- 5})$ $K_"sp" (=2.4 x 10^-5)$ > $Q_{sp} (1.6 x 10^{- 5})$ $Q_"sp"(1.6x10^-5)$

Explanation:

$20 m l (0.01 M C a C l_{2}) + 20 m l (0.008 M N a_{2} S O_{4})$ $20ml(0.01MCaCl_2) + 20ml(0.008MNa_2SO_4)$
=> $(0.02) (0.01) mole C a C l_{2}$ $(0.02)(0.01)"mole" CaCl_2$ + $(0.02) (0.008) mole N a_{2} S O_{4}$ $(0.02)(0.008)"mole" Na_2SO_4$
=> $0.0002 mole C a C l_{2}$ $0.0002"mole" CaCl_2$ + $0.00016 mole N a_{2} S O_{4}$ $0.00016"mole"Na_2SO_4$

=> $N a_{2} S O_{4}$ $Na_2SO_4$ is the limiting reagent => $1.6 x 10^{- 4}$ $1.6x10^-4"$ mole $C a S O_{4}$ $CaSO_4$

$[C a S O_{4}]$ $[CaSO_4]$ = $1.6 x 10^{- 4}$ $1.6x10^-4"$ $mole C a S O_{4}$ $"mole" CaSO_4$ / $0.040L Solution$ $"0.040L Solution"$ = $0.004 M$ $0.004M$

0.004M $C a S O_{4} ⇌$ $CaSO_4 rightleftharpoons$ +0.004M $(C a^{+ 2})$ $(Ca^(+2))$ + 0.004M $(S O_{4}^{2 -})$ $(SO_4^(2-))$

$Q_{s p} = [C a^{+ 3}] [S O_{4}^{2 -}] = (0.004 M) (0.004 M)$ $Q_(sp) = [Ca^(+3)][SO_4^(2-)] = (0.004M)(0.004M)$ = $1.6 x 10^{- 5}$ $1.6x10^-5$

$K_{s p} = 2.4 x 10^{- 5} > Q_{s p} = 1.6 x 10^{- 5}$ $K_(sp)=2.4x10^-5 > Q_(sp) = 1.6x10^-5$ => No Ppt will form.

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Question #a4cd9

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