Page 92 - E-BOOK KIMIA ANALITIK

=2,60

pCa pada titik ekivalen

50,0 × 0,00500
C 2− =
CaY
50,0 + 25,0
−3
= 3,33 × 10 M
2+
[Ca ] = c
T
2+
[CaY 2− ] = 0,00333 − [Ca ] ≊ 0,00333M
[CaY 2− ] 0,00333
10
= = 1,75 × 10
2+
2+ 2
[Ca ]c T [Ca ]
−7
2+
[Ca ] = √ 0,00333 = 4,36 × 10 M
1,75×10 10
−7
p Ca = −log4,36 × 10
=6,36

pCa setelah titik ekivalen

50,0 × 0,00500
C CaY 2− = 50,0 + 35,0

−3
= 2,94 × 10 M
35,0 × 0,0100 − 50,0 × 0,00500
C EDTA =
85
−3
= 1,18 × 10 M
−3
2+
[CaY 2− ] = 2,94 × 10 −3 − [Ca ] ≊ 2,94 × 10
−3
2+
c = 1,18 × 10 −3 + [Ca ] ≊ 1,18 × 10 M
T
2,94 × 10 −3
K′ =
CaY
2+
[Ca ] × 1,18 × 10 −3
10
= 1,75 × 10
2,94 × 10 −3
2+
[Ca ] =
1,18 × 10 −3 × 1,75 × 10 10
= 1,42 × 10 −10

p Ca = −log1,42 × 10 −10 = 9,85

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