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Ideal Gas Law

The pH of p...

Question

The $p H$ of pure water at $25^{o} C$ and $60^{o} C$ are $7$ and $6.5$ respectively. $H C l$ gas is passed through water at $25^{o} C$ till the resulting $1$ litre solution which aquires a $p H$ of $3$ . Now $4 \times 10^{- 3}$ mole of $N a C N$ are added into this solution. Also a fresh $0.1 M$ $H C N$ solution has $p H$ $5.1936$ . Now in the one part of solution obtained after addition of $N a C N$ , one milli mole of $N a O H$ are added and in the second part of this solution $0.5$ miili mole of $H C l$ are added.

The dissociation constant of $H C N$ is:

4.1 \times 10^{- 10}

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4.1 \times 10^{- 6}

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4.1 \times 10^{- 3}

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4.1 \times 10^{- 8}

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Solution

The correct option is A $4.1 \times 10^{- 10}$
$H C N 1 1 - α ⇌ H^{+} 0 α + {C N}^{-} 0 α$
Also,
$[H^{+}] = c \cdot α = c \sqrt{\frac{K_{a}}{c}} = \sqrt{K_{a} c}$
$- log H^{+} = - \frac{1}{2} [log K_{a} + log c]$
$5.1936 = - \frac{1}{2} [log K_{a} + log 0.1]$
$∴ K_{a} = 4.1 \times 10^{- 10}$

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Similar questions

Q. How many gram moles of

H C l

will be required to prepare one litre of a buffer solution (containing

N a C N

and

H C N

) of pH

8.5

using

0.10 g

formula mass of

N a C N

K_{a}

for

H C N = 4.1 \times 10^{- 10}

Q. How many g moles of HCl will be required to prepare one litre of buffer solution (containing NaCN and HCN) of pH= 8.5 using 0.01 gm formula mass of NaCN.

K_{a} (H C N) = 4.1 \times 10^{- 10}

How many gram moles of $H C l$ will be required to prepare one litre of a buffer solution (containing $N a C N$ and $H C N$ ) of $p H 8.5$ using $0.01 g r a m$ formula weight of $N a C N$ ?

$K_{a} (H C N)$ = $4.1 \times 10^{- 10}$

Q. How many mole of

H C N

will be required to prepare one litre buffer solution of

p H = 10.4

using

0.01

mol

N a C N

(Given:

K_{a} (H C N) = 4 \times 10^{- 10}

)

Q. A buffer of pH

8.50

is prepared from

0.02

mole of

K C N

. The desired volume of buffer solution is to be

1

litre. How will you make this buffer using

H C l

? What is the change after addition of

0.5 \times 10^{- 4}

mole

H C l

100 {c m}^{3}

and same amount of

N a O H

100 {c m}^{3}

of buffer?

(K_{a} H C N = 6.2 \times 10^{- 10})

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The correct option is A 4.1×10−10HCN11−α⇌H+0α+CN−0αAlso,[H+]=c⋅α=c√Kac=√Kac−logH+=−12[logKa+logc]5.1936=−12[logKa+log0.1]∴Ka=4.1×10−10