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Electrolysis and Electrolytes

Ka for hcn ...

Question

Ka for hcn is 5\ast10^{-10}.For maintaining a constant ph of 9, the volume of 5M kcn solution required to be added to 10ml of 2M hcn solution is A.2ml B.5ml C.3ml D.4ml

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Solution

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K_{a}

H C N

5 \times 10^{- 10}

25^{o} C

p H

9

5 M

K C N

10 m L

2 M

H C N

K_{a}

H C N

5 \times 10^{- 10}

25^{o} C

p H = 9

5 M

K C N

10 m l

2 M

H C N

{log}_{10} 5 = 0.69

(- 0.3010) = 0.5

K_{a}

5 \times 10^{- 10}

25^{0} C

K_{a}

5 \times 10^{- 10}

(log 2 = 0.3)

K_{a}

5 \times 10^{- 10}

25^{\circ} C

5 M K C N

10 m L

2 M H C N

(log 2 = 0.3 a n d l o g 5 = 0.69)

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