How much AgCN would precipitate out?

Question

HELP!!!!!!!!!!

enough AgCN is added to 1L of water to make a saturated solution. if 2L of a .046M solution of NaCN is added, how much AgCN would precipitate out? The Ksp for AgCN is 1.2*10^-16

HELP!! i have no clue how to do this problem...thanks

Answer 1

AgCN --> Ag+  &  CN-

K = [Ag+]  [CN-]

1.2e-16 =  [x]  [x]

x = amount of AgCN that went into solution as Ag+ & CN- is 1.095 e-8 molar in distilled water

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but when Cn- level has been raised to 0.046 molar

K = [Ag+]  [ CN-]

1.2e-16 =  [x]  [0.046]

x = amount of AgCN that went into solution as Ag+ & CN- is 2.609e-15 molar in 0.046M NaCN

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originally the 2 litres @ 1.095 e-8 mol/Litre had = 2.2 e-8 moles AgCN

but now the 2 litres @ 2.609e-15 has = 5 e-15 moles AgCN

the difference tells that virtually of the 2,2e-8 moles of AgCN has ppt'd out

2.2e-8 @ 133.89 g/mol =  2.9e-6 grams has ppt'd out

your answer is: 2.2e-8 moles , 2.9e-6 grams, of AgCN has ppt'd out