Question

Given `K_b=2.22xx10^-11` for nitrite ion, what is `pH` for a solution of `NaNO_2(aq)` that is `4.5xx10^-4*mol*L^-1` with respect to the salt?

Answer 1

`"pH" = 8.5`

Explanation:

The idea here is that sodium nitrite will dissociate completely in aqueous solution to produce sodium cations and nitrite anions, `"NO"_2^(-)`.

The nitrite anions will react with water to form nitrous acid, `"HNO"_2`, and produce hydroxide anions, so right from the start, you should expect the `"pH"` of the solution to be `>7`.

`"NO"_ (2(aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HNO"_ (2(aq)) + "OH"_ ((aq))^(-)`

For an aqueous solution at `25^@"C"`, the base dissociation constant, `K_b`, is equal to

`K_b = (1 * 10^(-14))/K_a`

with `K_a` being the acid dissociation constant of the base's conjugate acid.

The problem provides you with the acid dissociation constant of nitrous acid, so you can say that the nitrite anion will have

`K_b = (1 * 10^(-14))/(4.5 * 10^(-4)) = 2.22 * 10^(-11)`

By definition, the base dissociation constant is equal to

`K_b = (["HNO"_2] * ["OH"^(-)])/(["NO"_2^(-)])`

Now, notice that every more of nitrite anions that react produces `1` mole of nitrous acid and `1` mole of hydroxide anions.

This means that if you take `x` `"M"` to be the concentration of nitrite anions that react, you can say that, at equilibrium, the solution will contain

`["HNO"_2] = xcolor(white)(.)"M"`

`["OH"^(-)] = xcolor(white)(.)"M"`

The solution will also contain

`["NO"_2^(-)] = "0.5 M" - xcolor(white)(.)"M"`

This basically means that in order for the reaction to produce `x` `"M"` of nitrous acid and of hydroxide anions, it must consume `x` #"M"3 of nitrite anions.

Plug this into the expression of the base dissociation constant to get

`K_b = (x * x)/(0.5 - x)`

`2.22 * 10^(-11) = x^2/(0.5 - x)`

Now, because the value of the base dissociation constant is so small compared to the initial concentration of the nitrite anions, you can say that

`0.5 - x ~~ 0.5`

This means that you have

`2.22 * 10^(-11) = x^2/0.5`

which gets you

`x = sqrt(0.5 * 2.22 * 10^(-11)) = 3.33 * 10^(-6)`

The resulting solution will have

`["OH"^(-)] = 3.33 * 10^(-6)color(white)(.)"M"`

which means that its #"pH"# will be equal to

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

`"pH" = 14 - [-log(3.33 * 10^(-6))]`

`color(darkgreen)(ul(color(black)("pH" = 8.5)))`

The answer is rounded to one decimal place, the number of sig figs** you have for the concentration of the nitrite anions.

As predicted, the `"pH"` of the solution is indeed `>7`.

SIDE NOTE If you want, you can redo the calculations by taking into account the fact that the solution already contains a small concentration of hydroxide anions produced by the auto-ionization of water.

In that case, you would have

`["OH"^(-)] = (1 * 10^(-7) + x)color(white)(.)"M"`

`["HNO"_2] = xcolor(white)(.)"M"`

`["NO"_2^(-)] = (0.5 - x)color(white)(.)"M"`

with

`["OH"^(-)] = 3.28 * 10^(-6)color(white)(.)"M"`

Once again, you will have

`"pH" = 8.516 ~~ 8.5 ->` rounded to one decimal place