al→alcl3→al(oh)3→al2o3→na[al(oh)4]→al2(so4)3→al(oh)3→alcl3→na[al(oh)4]
1) 2al + 3cl2 = 2alcl3
al (0) - 3e > al (3+) процесс окисления, al восстановитель | 2|
cl2 (0) + 2e > 2cl (-1) процесс восстановления, сl2 окислиьтель | 3|
2) alcl3+ 3h2o (гор.) = al(oh)3↓ + 3hcl
al (3+)+ + 3h2o = al(oh)3↓ + 3h +3cl
al (3+) + 3h2o= al(oh)3↓
3)2al(oh)3 = al2o3 + 3h2o (при температуре)
4) 2naoh(конц., гор.) + 3h2o + al2o3 = 2na[al(oh)4]
5) такой реакции даже на сайте не нашла, даже в профильном учебнике не видела такого
6) al2(so4)3 + 6naoh (разб.) = 2al(oh)3↓ + 3na2so4
2al (3+) + 3so4 (2-) + 6na(+) + 6 oh = 2al(oh)3↓ + 6na(+) + 3so4 (2-)
2al (3+) + 6 oh = 2al(oh)3↓
7)al(oh)3 + 3hcl (разб.) = alcl3 + 3h2o
8) 4naoh(конц.) + alcl3 = na[al(oh)4] + 3nacl
There is a formula to calculate pH of buffer made of weak acid (here it's NaH2PO4) and its salt with strong base (Na2HPO4):
pH = pKa+lg CM Na2HPO4/CM NaH2PO4;
let's take that 450 ml of buffer is sum of volumes of Na2HPO4 and NaH2PO4 solutions (V1 ml - volume of Na2HPO4 sol. and V2 ml - volume of NaH2PO4 sol.), i.e. V1+V2 = 450 ml = 0.45 l. (1);
as CM = n/V and CM both of salts = 1 mol/l, so we have following: CM Na2HPO4 = V1/0.45 = 2.222*V1 and CM NaH2PO4 = V2/0.45 = 2.222*V2;
lg 2.222*V1/2.222*V2 = pH-pKa = 6.4-6.8 = -0.4, so 2.222*V1/2.222*V2 = 10^-0.8 = 0.3981;
as V2 = 0.45-V1 (see (1) above), we get 2.222*V1/2.222*(0.45-V1) = 0.3981, so V1 = 0.128 l. or 128 ml;
V2 = 0.45-0.128 = 0.322 l. or 322 ml;
Volume of Na2HPO4 sol. is 128 ml.;
Volume of NaH2PO4 sol. is 322 ml.
