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.
