NH4Cl = sale acido
In seguito ad idrolisi acida:
[H+] = 10^-5.13 = radice quadrata(Cs * Kw/Kb)
Cs = (10^-5.13)^2 * 1.8*10^-5 / 10^-14 = 0.1 M

pOH = 14 - 13 = 1
[OH-] = 10^-1 = 0.1
Ba(OH)2 => Ba2+ + 2 OH-
Ba(OH)2 = 0.1/2 = 0.05 M

500ml NH4Cl 0.1M
+
250ml Ba(OH)2 0.05M
=
750ml di
NH4Cl = 0.1M * 500/750 = 0.0667 M
e
Ba(OH)2 = 0.05M * 250/750 = 0.0167 M

2NH4Cl + Ba(OH)2 ---------> 2NH3 + 2 H2O + BaCl2
0.0667___0.0167_____________________________
0.0667-2x_0.0167-x_________2x____/_______/_____
con
0.0167 - x = 0
x = 0.0167
NH4Cl = 0.0667 - 2(0.0167) = 0.0333M
e NH3 = 2(0.0167) = 0.0333M
=
tampone ammoniacale:
[OH-] = Kb*[NH3]/[NH4+] = 1.8*10^-5 * 0.0333/0.0333 = 1.8*10^-5
pOH = 4.74
pH = 14 - 4.74 = 9.26