# Question 917b3

Jun 22, 2017

$\text{pH} = 11.20$

#### Explanation:

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

${\text{CN"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCN"_ ((aq)) + "OH}}_{\left(a q\right)}^{-}$

So even without doing any calculations, you should expect the $\text{pH}$ of the solution to come out $> 7$.

Now, look up the acid dissociation constant for hydrocyanic acid

${K}_{a} = 6.17 \cdot {10}^{- 10}$

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

${K}_{a} \cdot {K}_{b} = {K}_{W} = {10}^{- 14}$

to calculate the base dissociation constant of the cyanide anion.

${K}_{b} = \frac{{10}^{- 14}}{6.17 \cdot {10}^{- 10}} = 1.62 \cdot {10}^{- 5}$

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 $x$ $\text{M}$ of hydrocyanide acid and of hydroxide anions, you need to consume $x$ $\text{M}$ of cyanide anions.

By definition, the base dissociation constant is equal to

${K}_{b} = \left(\left[{\text{HCN"] * ["OH"^(-)])/(["CN}}^{-}\right]\right)$

In your case, this will be equal to

${K}_{b} = \frac{x \cdot x}{0.154 - x}$

$1.62 \cdot {10}^{- 5} = {x}^{2} / \left(0.154 - x\right)$

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

$0.154 - x \approx 0.154$

This means that you have

$1.62 \cdot {10}^{- 5} = {x}^{2} / 0.154$

Solve for $x$ to find

$x = \sqrt{0.154 \cdot 1.62 \cdot {10}^{- 5}} = 1.58 \cdot {10}^{- 3}$

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

$\left[{\text{OH}}^{-}\right] = 1.58 \cdot {10}^{- 3}$ $\text{M}$

An aqueous solution at room temperature has

$\text{pH + pOH = 14}$

so you can say that the $\text{pH}$ of the solution is equal to

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

$\text{pH} = 14 - \left[- \log \left(1.58 \cdot {10}^{- 3}\right)\right]$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{pH} = 11.20}}}$

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

$\text{pH} = 11.197$

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

As predicted, the $\text{pH}$ of the solution is $> 7$, which is consistent with the fact that the cyanide anion acts as a weak base in aqueous solution.