Напишите молекулярные уравнения реакций, протекающих в растворах, и запишите соответствующие им ионные уравнения1) Гидроксид калия + хлорид железа (III)2) Карбонат натрия + серная кислота​

3Fe(2+) + 6Na(+) + 2PO4(3-) + 3 SO4(2-) = Fe3(PO4)2 + 6Na(+) + 3SO4(2-)

3Fe(2+) + 2PO4(3-) = Fe3(PO4)2

 

Rb2SO3 + H2O = RbOH + RbHSO3

2Rb(+) + SO3(2-) + H2O <=> 2Rb(+) + HSO3(-) + OH(-)

SO3(2-) + H2O <=> HSO3(-) + OH(-)

Rb2SO3 = 2Rb(+) + SO3(2-) (electrolites and dissociation)

Hydrolysis of weak acid and strong base

OH(-) 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)2

Ca3(PO4)2 + 2H2O = 3Ca(2+) + 2HPO4(2-) + 2OH(-)

OH(-) is formed in process ==> pH > 7 (basic solution)

Ca(2+) and PO4(3-) - electrolites

диссоциацию солей на ионы.

Ca3(PO4)2 - weak electrolite

Ca3(PO4)2 <==> 3Ca(2+) + 2PO4(3-),  Ks = 1*10^-25 , T=298K

Hydrolysis of weak acid and strong base

Full hydrolysis (3 steps) Ca3(PO4)2 + 6H2O = 3Ca(OH)2 + 2H3PO4

2CaCO3 + 2H2O = Ca(HCO3)2 + Ca(OH)2

2CaCO3 + 2H2O = 2Ca(2+) + 2OH(-) + 2HCO3(-)

OH(-) is formed in process ==> pH > 7 (basic solution)

Ca(2+) and CO3(2-) - electrolites

CaCO3 - weak electrolite

CaCO3 <==> Ca(2+) + CO3(2-) , Ks = 4,9*10^-9 , T=298K

Hydrolysis of weak acid and strong base

Full hydrolysis (2 steps) CaCO3 +2H2O = Ca(OH)2 + "H2CO3"

Kb (NH3) = 2*10^-5

==> NH3 + H(+) = NH4(+)

[ ]-square brackets = molar concentration  mol/l

 [NH4(+)] = 2*10^-5 * [NH3]

==> [NH3] >>> [NH4(+)]

Ka Cl-CH2-COOH ~ 1*10^-2

Cl-CH2-COOH <==> Cl-CH2-COO(-) + H(+)

[Cl-CH2-COO(-)]<<[Cl-CH2-COOH]

Cl-CH2-COO(-) + NH4(+) <==> Cl-CH2-COONH4 (Salt)

speed of this reaction = k1[RCOO(-)][NH4(+)]

 

Cl-CH2-COOH + NH3 = NH2-CH2-COOH + HCl <==>(NH3(+)-CH2-COOH)Cl(-)

Sn2 reaction==> speed = k2[Cl-CH2-COOH][NH3]

If k1 > 10^7 k2 ==>both reactions goes with some speed

but more realistic version that speed of reaction2 >> speed of reaction1

==> NH2-CH2-COOH will be formed faster

==>NH2-CH2-COOH

Experimentally "...прилейте раствор HCl, затем обработайте щелочью, если образовывалась соль аммония, то выделится аммиак..."

 

Cl-CH2-COONH4 + HCl = Cl-CH2-COOH + NH4Cl

NH4Cl + OH(-) = NH3 + H2O + Cl(-)

 

NH2-CH2-COOH + HCl = (NH3(+)-CH2-COOH)Cl(-)

(NH3(+)-CH2-COOH)Cl(-) + OH(-) = NH2-CH2-COOH + H2O + Cl(-)