Question

The pH of 0.01 M NH4CN solution is (Ka(HCN) = 5 × 10–10, Kb(NH3) = 2 × 10–5) (log 5 = 0.7, log 2 = 0.3)

Answer 1

Answer:

pH=7+

2(pka−pkb)

1

pka=−logka

⇒−log(6.2×10

−10

)

⇒10−0.79=9.21

pkb=−logkb

⇒−log(1.6×10

−5

)

⇒5−0.20=4.8

7+

2

1

[9.21−4.8]

7+1/2[4.41]=9.21.

Explanation:

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