Question

Briefly review the reagents involved in the procedures for test tubes #2 and 3 (see pp. 8-3 and 8-4 in your lab manual.). Eac
Part 1- Could someone tell me which of these are soluble and insoluable.
Part 2- How do you write a net ionic equation for the precipitate?
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Answer #1
Anion Compound Test tube 2 (AgNO3) Test tube 3 (BaCl2) Overall molecular equation Net ionic equation
NaCH3COO

CH3COOAg (soluble)

(CH3COO)2Ba (soluble)

NaCH3COO + AgNO3 = NaNO3 + CH3COOAg

2 NaCH3COO + BaCl2 = 2 NaCl + (CH3COO)2Ba

CH3COO- (aq) + Ag+ (aq) = CH3COOAg (s)

2 CH3COO- (aq) + Ba2+ (aq) = (CH3COO)2Ba (s)

Na2CO3 Ag2CO3 (insoluble) BaCO3 (insoluble)

Na2CO3 + 2AgNO3 = 2 NaNO3 + Ag2CO3

Na2CO3 + BaCl2 = 2 NaCl + BaCO3

2Ag+ (aq) + CO32- (aq) = Ag2CO3 (s)

Ba2+ (aq) + CO32-(aq) = BaCO3 (s)

NaF

AgF

(soluble)

BaF2 (insoluble)

NaF + AgNO3 = NaNO3 + AgF

2NaF + BaCl2 = 2NaCl + BaF2

Ag+(aq) + F- (aq) = AgF (s)

Ba2+ (aq) + 2F- (aq) = BaF2 (s)

NaCl

AgCl

(insoluble)

No precipitate

NaCl + AgNO3 = NaNO3 + AgCl

2NaCl + BaCl2 = No reaction

Ag+(aq) + Cl- (aq) = AgCl (s)

NaBr AgBr (insoluble)

BaBr2

(soluble)

NaBr + AgNO3 = NaNO3 + AgBr

2 NaBr + BaCl2 = 2 NaCl + BaBr2

Ag+(aq) + Br- (aq) = AgBr (s)

Ba2+ (aq) + 2Br- (aq) = BaBr2 (s)

NaI

AgI

(insoluble)

BaI2

(soluble)

NaI + AgNO3 = NaNO3 + AgI

2 NaI + BaCl2 = 2 NaCl + BaI2

Ag+(aq) + I- (aq) = AgI (s)

Ba2+ (aq) + 2 I- (aq) = BaI2 (s)

NaNO3 No precipitate Ba(NO3)2 (soluble)

NaNO3 + AgNO3 = No reaction

2 NaNO3 + BaCl2 = 2 NaCl + Ba(NO3)2

Ba2+ (aq) + 2NO3- (aq) = Ba(NO3)2 (s)

NaSCN AgSCN (insoluble) BaSCN (soluble)

NaSCN + AgNO3 = NaNO3 + AgSCN

2 NaSCN + BaCl2 = 2 NaCl + BaSCN

Ag+ (aq) + SCN- (aq) = AgSCN (s)

Ba2+ (aq) + 2 SCN- (aq) = BaSCN (s)

Na2SO4 Ag2SO4 (insoluble) BaSO4 (insoluble)

Na2SO4 + 2AgNO3 = 2NaNO3 + Ag2SO4

Na2SO4 + BaCl2 = 2 NaCl + BaSO4

2Ag+(aq) + SO42- (aq) = Ag2SO4 (s)

Ba2+(aq) + SO42- (aq) = BaSO4 (s)

Na3PO4

Ag3PO4

(insoluble)

Ba3(PO4)2 (insoluble)

Na3PO4 + 3AgNO3 = 3NaNO3 + Ag3PO4

2Na3PO4 + 3BaCl2 = 6 NaCl + 2Ba3(PO4)2

3Ag+ (aq) + PO43- (aq) = Ag3PO4 (s)

3Ba2+ (aq) + PO43- (aq) = 2Ba3(PO4)2 (s)

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