Alright, this is a simple I.C.E. type of equation. (I.C.E. means setting up a little table where you list the Initial, Change, and Equillibrium concentrations of all species involved in the reaction). HCN is a weak acid, so when a cyanide salt is placed in water the solution becomes basic in nature. (Because the CN- ions will grab an H+ from water to form HCN and leave OH- ions in solution). Therefore, we will need to convert the Ka of HCN into the Kb of CN- to do the equation properly. (Remember that Kb = Kw/Ka). Since you know the Ka of HCN, you can find the Kb of CN- by using (1.0x10^-14)/(4.0x10^-10). This will give you the Kb of CN-.
Now the equation you will need to use is the following:
Kb(CN-) = ([HCN][OH-])/[CN-]
Since you know the initial concentration of CN-, you can figure out the equillibrium concentrations of OH- and HCN.
Initial:
HCN = 0
OH- = 0
CN- = 1.75 M
Change:
HCN = x
OH- = x
CN- = -x
Equillibrium:
HCN = x
OH- = x
CN- = 1.75 - x
Using the equation written out above, you solve for x. Once you get the value of x, you'll then have the value of the OH- ion concentration at equillibrium. If you take the -log of that concentration, you will have yourself the pOH of the solution.
Since pOH + pH = 14, you can quite easily determine the pH of the solution.
Let me know if you need further help with this.