# Question #bb890

##### 1 Answer

#### Explanation:

You can solve this problem without knowing the **acid dissociation constant**, **here**.

#K_a = 6.17 * 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

#"HCN"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "CN"_ ((aq))^(-)#

will lie heavily **to the left**.

Now, if you take **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# #"M"# of hydronium cations and#x# #"M"# of cyanide anions, it must consume#x# #"M"# of hydrocyanide acid.

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

#K_a = (["H"_3"O"^(+)] * ["CN"^(-)])/(["HCN"])#

In your case, this is equivalent to

#6.17 * 10^(-10) = (x * x)/(1 * 10^(-2) - x)#

#6.17 * 10^(-10) = x^2/(1 * 10^(-2) - x)#

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 * 10^(-2) - x ~~ 1 * 10^(-2)#

This means that you have

#6.17 * 10^(-10) = x^2/(1 * 10^(-2))#

which gets you

#x = sqrt(6.17 * 10^(-10) * 1 * 10^(-2)) = 2.484 * 10^(-6)#

So you can say that only a concentration of **will ionize** to produce hydronium cations and cyanide anions.

This means that the **degree of ionization**, or

#alpha = (2.484 * 10^(-6) color(red)(cancel(color(black)("M"))))/(1 * 10^(-2) color(red)(cancel(color(black)("M")))) = 0.02484 ~~ color(darkgreen)(ul(color(black)(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

**SIDE NOTE** *When dealing with such small concentrations of hydronium cations, you should account for the fact that pure water at #25^@"C"# already contains*

*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 = ((1 * 10^(-7) + x) * x)/(1 * 10^(-2) - x)#

*Using the same approximation as before, you can say that*

#x^2 + 1 * 10^(-7) * x - 6.17 * 10^(-12) = 0#

*This time, you would end up with*

#x = 2.434 * 10^(-6)#

*and a degree of ionization of*

#alpha = (2.434 * 10^(-6) color(red)(cancel(color(black)("M"))))/(1 * 10^(-2) color(red)(cancel(color(black)("M")))) = 0.02434 ~~ 0.02#

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