Grupa 4 cationi

Question 1

calciu + carbonat de amoniu

Answer 1

Ca2+ + (NH4)2CO3 = CaCO3↓ + 2NH4
pp alb amorf – pp alb
usor sol in CH3COOH
CaCO3 + 2CH3COOH = (CH3COO)2Ca + CO2↑+ H2O

Question 2

calciu+ oxalat de amoniu

Answer 2

Ca2+ + (COONH4)2 = (COO)2Ca↓ + 2NH4
pp alb cristalin - sol in acizi minerali
Ca(COO)2 + H2SO4 = CaSO4 + (COOH)2

Question 3

calciu + acid sulfuric

Answer 3

Ca2+ + H2SO4 = CaSO4↓ + 2H+
pp alb cristalin - sol in:
CaSO4 + (NH4)2SO4 = (NH4)2Ca(SO4)2

Question 4

calciu+ murexid

Answer 4

sarea de amoniu a acidului purpuric formează cu ionul Ca2+ la un pH>12, un complex solubil, colorat în violet - purpurat de Ca

Question 5

calciu + ferocianura de potasiu II

Answer 5

Ca2+ + K4[Fe(CN)6] + 2NH4Cl = (NH4)2Ca[Fe(CN)6]↓ + 4K+ + 2Cl‾
pp alb

Question 6

strontiu+ carbonat de amoniu

Answer 6

Sr2+ + (NH4)2CO3 = SrCO3↓ + 2NH4
SrCO3 + 2CH3COOH = (CH3COO)2Sr + CO2↑ + H2O

Question 7

strontiu+acid sulfuric

Answer 7

Sr2+ + H2SO4 = SrSO4↓ + 2H+ -reactia are loc mai repede la cald
pp alb
SrSO4 + 2HCl = SrCl2 + H2SO4

Question 8

strontiu+sulfat de amoniu

Answer 8

Sr2+ + (NH4)2SO4 = SrSO4↓ + 2NH4
pp solubil in HCl conc, la cald

Question 9

strontiu+cromat de potasiu

Answer 9

Sr2+ + K2CrO4 = SrCrO4↓ + 2K+
NUMAI in sol conc,
NU are loc in prez de CH3COOH

Question 10

bariu+carbonat de amoniu

Answer 10

Ba2+ + (NH4)2CO3 = BaCO3↓ + 2NH4
pp alb cristalin - sol in acizi
BaCO3 + 2CH3COOH = (CH3COO)2Ba + CO2↑ + H2O

Question 11

bariu+acid sulfuric

Answer 11

Ba2+ + H2SO4 = BaSO4↓ + 2H+
pp alb cristalin

Question 12

bariu+cromat de potasiu

Answer 12

Ba2+ + K2CrO4 = BaCrO4↓ + 2K+
pp galben sol in acizi minerali
BaCrO4 + HCl = BaCl2+ H2CrO4

Question 13

bariu+ rodizonat de sodiu

Answer 13

Ba2+ + C6O6Na2 = C6O6Ba + 2Na+
pp rosu
r se poate executa si pe hf
se produce in sol neutra/slab acida

Question 14

cationi gr 4+ reactiv de grupa

Answer 14

Ca2+ + (NH4)2CO3 = CaCO3↓ + 2NH4
pp alb amorf – pp alb
usor sol in CH3COOH
CaCO3 + 2CH3COOH = (CH3COO)2Ca + CO2↑+ H2O
Sr2+ + (NH4)2CO3 = SrCO3↓ + 2NH4
SrCO3 + 2CH3COOH = (CH3COO)2Sr + CO2↑ + H2O
Ba2+ + (NH4)2CO3 = BaCO3↓ + 2NH4
pp alb cristalin - sol in acizi
BaCO3 + 2CH3COOH = (CH3COO)2Ba + CO2↑ + H2O