Question

The pH of a 2M solution of NaOH is...? (please include the complete solution of the problem)

Answer 1

`"pH" = 14.3`

Explanation:

For starters, you should know that an aqueous solution at room temperature has

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

This means that you express the `"pH"` of the solution in terms of its `"pOH"` by writing

`"pH" = 14 - "pOH"`

Now, the `"pOH"` of the solution is defined as the negative log base `10` of the concentration of hydroxide anions

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

This means that the `"pH"` of the solution can be expressed as

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

`"pH" = 14 + log(["OH"^(-)])" "color(darkorange)("(*)")`

As you know, sodium hydroxide is a strong base, which implies that it ionizes completely in aqueous solution to produce sodium cations and hydroxide anions.

You can thus say that

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

Since every `1` mole of sodium hydroxide that is dissolved in water produces `1` mole of hydroxide anions, your solution will have

`["OH"^(-)] = ["NaOH"]`

In your case, this is equal to

`["OH"^(-)] = "2 M"`

Plug this into equation `color(orange)("(*)")` and calculate the `"pH"` of the solution

`"pH" = 14 + log(2) = color(darkgreen)(ul(color(black)(14.3)))`

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

As a final note, don't get confused by the fact that you have

`"pH" > 14`

This is what you'll always get for basic solutions that have

`["OH"^(-)] > 1`

and it means that `"pOH" < 0`.