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What amount of HCl will be required to p...

What amount of `HCl` will be required to prepare one litre of a buffer solution of `pH 10.4` using `0.01` mole of `NaCN`? Given `K_(ion)(HCN)=4.1xx10^(-10)`.

A

`8.55xx 10^(-3)"mole"`

B

`8.65xx 10^(-3)"mole"`

C

`8.75xx 10^(-3)"mole"`

D

`9.09xx 10^(-3)"mole"`

Text Solution

Verified by Experts

The correct Answer is:
D
The addition of HCl converts NaCN into HCN. Let x be the amount of HCl added. We will have.
[NaCN] = (0.01 - x)
[HCN] = x
Substituting these values along with pH and Ka in the expression.
`pH =-log K_(a)+log""(["Salt"])/(["Acid"])`
We get `10.4=-log[4 xx 10^(-10)]+log (0.01-x)/(x) or 10.4=9.4 +log""(0.01-x)/(x)`
`or log"" (0.1-x)/(x)=1`
`or (0.01-x)/(x)=10 rArr 11 x=10^(-2)`
`or x=9.09 xx 10^(-4)M`
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