Question #6d0d9

1 Answer
Nov 6, 2017

A: 50 ml
B:pH=2.87
C:pH=4.74
D:pH=8.72

Explanation:

A:
50 ml HAc 0.1M = 5mMol HAc (weak acid)
50 NaOH 0.1 M = 5 mMol NaOH (strong base)
Equivalence: 1:1, 5mMol:5 mMol => 50ml 0,.1 M NaOH

B:
pH=logKaF
=> pH=log1.850.1M
=> pH=log1.86
=> pH=log1.34103
=> pH=2.87

C:
pH=pKa+log(ηOHηHAinitηOH)

pH=pKa+log(2.535.032.53)

=>pH=pKa+log(2.532.53)
=>pH=pKa+log1
log1 = 0, so:
pH=pKa = logKa=4.74

D:
pH=14+logCaCbKw(Ca+Cb)Ka

pH=14+log0.10.111014(0.1+0.1)1.8105
pH=14+log110160.21.8105
pH=14+log110103.6=145.28
pH=8.72