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0,112 liter = mol x 22,4
mol = 0,112 = 0,005 mol
22,4
mol HCl = 0,02 liter x 0,15 M = 0,003 mol
NH₃ + HCl ---> NH₄Cl
m 0,005 0,003 -
r 0,003 0,003 0,003
s 0,002 - 0,003
[OH⁻] = kb x nb
gb
[OH⁻] = 10⁻⁵ x 0,002
0,003
[OH⁻] = 10⁻⁵ x 2
3
pOH = -log [OH⁻]
= 5 - (log 2-log3)
= 5 - log 2 + log 3
= 5 - 0,3 + 0,5
= 5,2
pH = pKw - pOH
= 14 - 5,2
= 8,8