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Q.

Match the conditions on left column with solution on right column.

Column – IColumn – II
A) Has maximum pH  at the end point when titrated against KOH P) CH3COOH(Ka=2×105)
B) Has minimum  pH at the end point when titrated with  KOHQ) HCN(Ka=5×1010)
C) Evolve maximum heat when treated with  NaOH R) HF(Ka=5×104)
D) Evolve equal amount of heat when titrated with strong acid or strong base S) NH3(Kb=2×105)

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a

A – R, B – Q, C – R, D – QS

b

A – Q, B – R, C – R, D – PS

c

A – P, B – R, C – R, D – PS

d

A – Q, B – R, C – R, D – RS

answer is B.

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Detailed Solution

Lower be the ka  value of acid, higher be the degree of hydrolysis of its salt of strong base. 
Stronger acids release maximum heat when titrated with strong base.
Equal values of ka&kb  for acid and base respectively will evolve same amount of heat and titration.
 

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Match the conditions on left column with solution on right column.Column – IColumn – IIA) Has maximum pH  at the end point when titrated against KOH P) CH3COOH(Ka=2×10−5)B) Has minimum  pH at the end point when titrated with  KOHQ) HCN(Ka=5×10−10)C) Evolve maximum heat when treated with  NaOH R) HF(Ka=5×10−4)D) Evolve equal amount of heat when titrated with strong acid or strong base S) NH3(Kb=2×10−5)