Homework Answers
pH = -log[H+]
pH = log 1/[H+]
● pH of 0.0045 M phenol solution
[H+] Concentration = 0.0045
pH = - log [H+]
pH = log 1/[H+]
pH = log 1/0.0045
pH = log 10000/45
pH = log 10000 - (log9×5)
pH = log 10000 - (log9 + log5)
pH = 4 - (0.954 + 0.699)
pH = 4 - 1.653
PH = 2.347
● pH of 0.0052M ammonia solution
pH = log 1/[H+]
pH = log 1/0.0052
pH = log 10000/52
pH = log 10000 - ( log13×4)
pH = log 10000 - (log13 + log 4)
pH = 4 - (1.114 + 0.602)
pH = 4 - 1.716
pH = 2.284
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Determine the ammonia concentration of an aqueous solution that
has a pH of 11.30. The equation for the dissociation of
NH3 (Kb = 1.8 × 10-5) is
below:
Determine the ammonia concentration of an aqueous solution that
has a pH of 11.30. The equation for the dissociation of
NH3 (Kb = 1.8 × 10-5) is
below:
a)9.0 × 10-3 mol L-1
b)2.7 mol L-1
c)0.22 mol L-1
d)2.0 × 10-3 mol L-1
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