Question #917b3

The idea here is that the cyanide anion, #"CN"^(-)#, will act as a weak base in aqueous solution, .e. it will react with water to form hydrocyanic acid, #"HCN"#, and hydroxide anions, #"OH"^(-)#.

#"CN"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCN"_ ((aq)) + "OH"_ ((aq))^(-)#

So even without doing any calculations, you should expect the #"pH"# of the solution to come out #>7#.

Now, look up the acid dissociation constant for hydrocyanic acid

#K_a = 6.17 * 10^(-10)#

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

Use the fact that an aqueous solution at room temperature has

#K_a * K_b = K_W = 10^(-14)#

to calculate the base dissociation constant of the cyanide anion.

#K_b = (10^(-14))/(6.17 * 10^(-10)) = 1.62 * 10^(-5)#

Now, notice that every mole of cyanide anions that picks up a proton from water forms #1# mole of hydrocyanic acid and #1# mole of hydroxide anions.

If you take #x# to be the equilibrium concentration of hydrocyanide acid and of hydroxide anions, you can say that

#["CN"^(-)] = ["CN"^(-)]_ "initial" - x#

This means that in order to have a concentration of #x# #"M"# of hydrocyanide acid and of hydroxide anions, you need to consume #x# #"M"# of cyanide anions.

By definition, the base dissociation constant is equal to

#K_b = (["HCN"] * ["OH"^(-)])/(["CN"^(-)])#

In your case, this will be equal to

#K_b = (x * x)/(0.154 - x)#

#1.62 * 10^(-5) = x^2/(0.154 - x)#

Now, notice that the base dissociation constant is considerably smaller than the initial concentration of the cyanide anions, so use the approximation

#0.154 - x ~~ 0.154#

This means that you have

#1.62 * 10^(-5) = x^2/0.154#

Solve for #x# to find

#x = sqrt(0.154 * 1.62 * 10^(-5)) = 1.58 * 10^(-3)#

Since #x# represents the equilibrium concentration of the hydroxide anions, you can say that you have

#["OH"^(-)] = 1.58 * 10^(-3)# #"M"#

An aqueous solution at room temperature has

#"pH + pOH = 14"#

so you can say that the #"pH"# of the solution is equal to

#"pH" = 14 - [-log(["OH"^(-)])]#

#"pH" = 14 - [-log(1.58 * 10^(-3))]#

#color(darkgreen)(ul(color(black)("pH" = 11.20)))#

I'll leave the answer rounded to two decimal places, but you can leave the answer rounded to three decimal places

#"pH" = 11.197#

because you have three sig figs for the initial concentration of the acid.

As predicted, the #"pH"# of the solution is #>7#, which is consistent with the fact that the cyanide anion acts as a weak base in aqueous solution.