How many gram moles of `HCI` will be required to prepare `1L` of buffer solution (containing `NaCN` and `HCI)` of `pH 8.5` using `0.01g` formula weight of `NaCN gt K_(HCN) = 4.1 xx 10^(-10)`.

Answer

A, C

Explanation

Let the number of moles of `HCI` required be `n`.
After adding `HCI`, the buffer is, formed as follows:
`{:(NaCN,HCIrarr,NaCI+,HCN),(0.01,n,0,0),((0.01-n),0,n,n):}`
`pH = pK_(a) +"log" (["Salt"])/(["Acid"])`
`= pK_(a) +"log" ([NaCN])/([HCN])`
or `8.5 =- log 4.1 xx 10^(-10) ((0.01-n))/(n)`
or `n~~8.85 xx 10^(-3)M`