Question

What is the pH of a `1.4 * 10^-2 M` NaOH solution?

Answer 1

`"pH" = 12.15`

Explanation:

Even before doing any calculations, you can say that since you're dealing with a strong base, the pH of the solution must be higher than `7`.

The higher the concentration of the base, the higher the pH will be.

In your case, you're dealing with a solution of sodium hydroxide, `"NaOH"`, a strong base that dissociates completely in aqueous solution to form sodium cations, `"Na"^(+)`, and hydroxide anions, `"OH"^(-)`

`"NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH"_text((aq])^(-)`

Notice that the salt dissociates in a `1:1` mole ratio with the hydroxide anions, you you can say that

`["OH"^(-)] = ["NaOH"] = 1.4 * 10^(-2)"M"`

Now, the pH of the solution is determined by the concentration of hydronium ions, `"H"_3"O"^(+)`. For aqueous solutions, the concentration of hydronium ions is related to the concentration of hydroxide ions by the ion product constant of water, `K_W`

`K_W = ["OH"^(-)] * ["H"_3"O"^(+)]`

At room temperature, you have

`K_W = 10^(-14)`

This means that the concentration of hydronium ions can be determined by using

`["H"_3"O"^(+)] = K_W/(["OH"^(-)])`

Plug in your values to get

`["H"_3"O"^(+)] = 10^(-14)/(1.4 * 10^(-2)) = 7.14 * 10^(-13)"M"`

The pH of the solution is equal to

`color(blue)("pH" = - log(["H"_3"O"^(+)])`

In your case,

`"pH" = -log(7.14 * 10^(-13)) = color(green)(12.15)`

As predicted, the pH is not only higher than `7`, but it is significantly higher than `7`.

Alternatively, you can use the pOH of the solution to find its pH. As you know,

`color(blue)("pOH" = - log(["OH"^(-)]))`

In your case,

`"pOH" = - log(1.4 * 10^(-2)) = 1.85`

You know that

`color(blue)("pH" + "pOH" = 14)`

and so, once again, you have

`"pH" = 14 - 1.85 = color(green)(12.15)`