FeSO4+Na3PO4=Fe3 (PO4) 2+NaSO4 (a) -RbSO3, Ca2 (PO4) 3,CaCO3. Для уравнения напишите схемы в полной и…

3Fe (2+)+6Na (+)+2PO4 (3-)+3 SO4 (2-)=Fe3 (PO4) 2+6Na (+)+3SO4 (2-) 3Fe (2+)+2PO4 (3-)=Fe3 (PO4) 2 Rb2SO3+H2O=RbOH+RbHSO32Rb (+)+SO3 (2-)+H2O < => 2Rb (+)+HSO3 (-)+OH (-) SO3 (2-)+H2O < => HSO3 (-)+OH (-) Rb2SO3=2Rb (+)+SO3 (2-) (electrolites and dissociation) Hydrolysis of weak acid and strong baseOH (-) is formed in process => pH > 7 (basic solution) «соль которая образуется в конце полного гидролиза."Full hydrolysis (2 steps) Rb2SO3+2H2O=2RbOH+"H2SO3"How you can see, salts doesn't formed in full hydrolysis. Ca3 (PO4) 2+2H2O=2Ca (HPO4)+Ca (OH) 2Ca3 (PO4) 2+2H2O=3Ca (2+)+2HPO4 (2-)+2OH (-) OH (-) is formed in process => pH > 7 (basic solution) Ca (2+) and PO4 (3-) — electrolitesдиссоциацию солей на ионы.Ca3 (PO4) 2 — weak electroliteCa3 (PO4) 2 < => 3Ca (2+)+2PO4 (3-) , Ks=1*10^-25 , T=298KHydrolysis of weak acid and strong baseFull hydrolysis (3 steps) Ca3 (PO4) 2+6H2O=3Ca (OH) 2+2H3PO42CaCO3+2H2O=Ca (HCO3) 2+Ca (OH) 22CaCO3+2H2O=2Ca (2+)+2OH (-)+2HCO3 (-) OH (-) is formed in process => pH > 7 (basic solution) Ca (2+) and CO3 (2-) — electrolitesCaCO3 — weak electroliteCaCO3 < => Ca (2+)+CO3 (2-) , Ks=4,9*10^-9 , T=298KHydrolysis of weak acid and strong baseFull hydrolysis (2 steps) CaCO3+2H2O=Ca (OH) 2+"H2CO3"