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Q.

20 mL of a weak monobasic acid (HA) requires 20 mL 0.2 M NaOH for complete titration. If pH of solution upon addition of 10 mL of this alkali to 25 mL of the above solution of HA is 5.8. The pKa of the weak acid is

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a

6.1

b

5.8

c

5.98

d

5.58

answer is C.

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Detailed Solution

meq. of acid = meq of base   20 × M = 20 × 0.2 = 4

Molarity of HA = 0.2 M

When 25ml,0.2M acid reacts with 10ml 0.2M NaOH

         HA    +      OH-             A-      +         H2O    5 (=25×.2)     2 (10×.2)              0      3(=5-2)              -                 2 mmole       pH = pKa + log   A-HA     5.8 =  pKa +  log 23      pKa = 5.98

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