Hydrocyanic acid is a weak acid that ionises:
HCN_((aq))rightleftharpoonsH_((aq))^(+)+CN_((aq))^-
For which:
K_a=([H_((aq))^+][CN_((aq))^-])/([HCN_((aq))])
These are equilibrium concentrations.
Rearranging gives:
[H_((aq))^+]=K_axx[[HCN_((aq))]]/[[CN_((aq))^-]]
We can use the data given to get the acid/salt ratio.
pH=8.5
:.log[H_((aq))^+]=-8.5
From which [H_((aq))^+]=3.16xx10^(-9)"mol/l"
Putting in the numbers rArr
3.16xx10^(-9)=4.1xx10^(-10)xx([HCN_((aq))])/([CN_((aq))^-])
:.([HCN_((aq))])/([CN_((aq))^-])=(3.16xx10^(-9))/(4.1xx10^(-10))=0.77
Because K_a is so small we can say that the equilibrium shown above lies well to the left.
We can therefore assume that if H^+ ions are added to a solution containing CN^- (nitrile) ions then they will form HCN and we can ignore the tiny amount that dissociate.
So if n moles of H^+ ions are added to a solution containing 0.01 moles of CN^- then n moles of HCN form and 0.01-n moles of CN^- remain.
Since the total volume is common to both HCN and CN^- we can write:
(n)/(0.01-n)=0.77
:.n=0.77(0.01-n)
:.n=0.0077-0.77n
:.n=0.0077/1.77=0.00435"mol"
This is the number of moles of HCl needed.
The question does not specify any concentration of HCl to be used so here's a possible recipe for the buffer:
Say we are given HCl of concentration 1"mol/l"
How much would we need?
c=n/v:.v=n/c=0.00435/1=0.00435"L"=4.35"ml"
The M_r of NaCN = 49.01
:.0.01"mol"=0.49"g"
You could dissolve this in a minimum amount of distilled water and transfer, with washings, to a 1 litre volumetric flask.
Then add 4.35"ml" of 1"M""" ""HCl from a graduated pipette and make up to the 1 litre mark.
