# Question bb890

Feb 8, 2018

$\alpha = 0.02$

#### Explanation:

You can solve this problem without knowing the acid dissociation constant, ${K}_{a}$, of hydrocyanic acid, which you can find listed here.

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

Now, hydrocyanic acid is a weak acid, which implies that it does not ionize completely in aqueous solution. In other words, not every molecule of hydrocyanic acid will actually donate its acidic proton to a water molecule to produce hydronium cations and cyanide anions.

In fact, the very small value of the acid dissociation constant tells you that the vast majority of molecules of hydrocyanic acid will not ionize in aqueous solution.

So the ionization equilibrium of hydrocyanic acid

${\text{HCN"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "CN}}_{\left(a q\right)}^{-}$

will lie heavily to the left.

Now, if you take $x$ $\text{M}$ to be the concentration of hydrocyanic acid that does ionize to produce hydronium cations and cyanide anions, you can say that, at equilibrium, the solution will contain

["H"_ 3"O"^(+)] = x quad "M"

["CN"^(-)] = x quad "M"

["HCN"] = (1 * 10^(-2) - x) quad "M"

This basically means that in order for the reaction to produce $x$ $\text{M}$ of hydronium cations and $x$ $\text{M}$ of cyanide anions, it must consume $x$ $\text{M}$ of hydrocyanide acid.

By definition, the acid dissociation constant that describes this equilibrium is equal to

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["CN"^(-)])/(["HCN}\right]\right)$

In your case, this is equivalent to

$6.17 \cdot {10}^{- 10} = \frac{x \cdot x}{1 \cdot {10}^{- 2} - x}$

$6.17 \cdot {10}^{- 10} = {x}^{2} / \left(1 \cdot {10}^{- 2} - x\right)$

Since the value of the acid dissociation constant is so small compared to the initial concentration of the acid, you can use the approximation

$1 \cdot {10}^{- 2} - x \approx 1 \cdot {10}^{- 2}$

This means that you have

$6.17 \cdot {10}^{- 10} = {x}^{2} / \left(1 \cdot {10}^{- 2}\right)$

which gets you

$x = \sqrt{6.17 \cdot {10}^{- 10} \cdot 1 \cdot {10}^{- 2}} = 2.484 \cdot {10}^{- 6}$

So you can say that only a concentration of $2.484 \cdot {10}^{- 6}$ $\text{M}$ out of an initial concentration of $1 8 {10}^{- 2}$ $\text{M}$ of hydrocyanic acid will ionize to produce hydronium cations and cyanide anions.

This means that the degree of ionization, or $\alpha$, is equal to

$\alpha = \left(2.484 \cdot {10}^{- 6} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{M"))))/(1 * 10^(-2) color(red)(cancel(color(black)("M}}}}\right) = 0.02484 \approx \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{0.02}}}$

You can also calculate the percent ionization of the acid

"% ionization" = (2.484 * 10^(-6) color(red)(cancel(color(black)("M"))))/(1 * 10^(-2) color(red)(cancel(color(black)("M")))) * 100% = 0.02484% ~~ color(darkgreen)(ul(color(black)(0.02%)))

The answer is rounded to one significant figure.

This value tells you that out of every $\text{10,000}$ molecules of hydrocyanic acid present in the solution, only $2$ will ionize.

$\textcolor{w h i t e}{\frac{a}{a}}$
SIDE NOTE When dealing with such small concentrations of hydronium cations, you should account for the fact that pure water at ${25}^{\circ} \text{C}$ already contains $1 \cdot {10}^{- 7}$ $\text{M}$ of hydronium cations produced by the auto-ionization of water.

So a more accurate approach would be to say that, at equilibrium, the solution contains

["H"_ 3"O"^(+)] = (1 * 10^(-7) + x) quad "M"

["CN"^(-)] = x quad "M"

["HCN"] = (1 * 10^(-2) - x) quad "M"#

The acid dissociation constant would be equal to

${K}_{a} = \frac{\left(1 \cdot {10}^{- 7} + x\right) \cdot x}{1 \cdot {10}^{- 2} - x}$

Using the same approximation as before, you can say that

${x}^{2} + 1 \cdot {10}^{- 7} \cdot x - 6.17 \cdot {10}^{- 12} = 0$

This time, you would end up with

$x = 2.434 \cdot {10}^{- 6}$

and a degree of ionization of

$\alpha = \left(2.434 \cdot {10}^{- 6} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{M"))))/(1 * 10^(-2) color(red)(cancel(color(black)("M}}}}\right) = 0.02434 \approx 0.02$

Once again, the answer must be rounded to one significant figure.