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Q.

Match the effect of addition of 0.1 M NaOH to 0.1 M, 50 mL H3 PO4 Ka1, Ka2, Ka3 are the I, II, III ionization constant of H3PO4 respectively.

 Column-I
(Vol. of NaOH added)
 Column-II
(pH of resultant solution)
A)150 mL of NaOHP)pH=pka2
B)75 mL of NaOHQ)pH=pKa1
C)25 mL of NaOHR)pH=7+pKa3+logc2
D)100 mL of NaOHS)pH=pka2+pka32
 ABCD
1.TPQ,R,SR
2.P,Q,R,SQP,Q,R,SQ
3.PPQS
4.RPQS

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a

2

b

1

c

 4

d

3

answer is D.

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Detailed Solution

H3PO4   +   NaOHNaH2PO4+H2O0.1×50        0.1×255m mole       2.5m mole  02.5                          2.5mmole2.5mmole
pH=pKa1
Further addition of 25ml NaOH, only NaH2PO4 (5 m moles) will be formed

Further addition of 25 ml NaOH (Total 75 ml),
NaH2PO4+NaOHNa2HPO4+H2O  5m moles 2.5 moles 02.5 2.5 mmoles 2.5m moles  pH=pKa2  
Further addition of 25 ml NaOH, only Na2 HPO4 (5 m moles) will be formed
pH=pKa2+PKa32
Further addition of 25 ml (Total 125 ml)
Na2HPO4+NaOHNa3PO4+H2O5m moles  2.5mmoles 02.5 2.5m moles  2.5mmolespH=pKa3
Further addition of 25 m l (Total 150 m l Na3PO4 salt will be formed
pH=7+pKa32+12logC

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