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Q.

100 mL of 0.1 M HCI is titrared by using 0.1 M NaOH, using phenolphthalein indicator. pH after 50 mL of NaOH has been added is

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a

7.0

b

-1.48

c

6.0

d

1.48

answer is D.

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Detailed Solution

0.1MHCl=100mL ∴ N1V1( acid )=10 0.1MNaOH=50mL N2V2 (base) =5  Thus, N1V1>N2V2 Resultant mixture is acidic  Resultant H+=N1V1−N2V2V1+V2=130 ∴ pH=−logH+=−log130=1.48
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