Question

Question #0f970

Answer 1

Here's what I got.

Explanation:

One way to tackle this problem would be to use the fact that an aqueous solution at `25^@"C"` has

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

and that

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

to express the `"pH"` of the solution in terms of the concentration of hydroxide anions.

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

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

Now, you know that the concentration of hydroxide anions is given by the number of moles of hydroxide anions present for every `"1 L" = 10^3 quad "mL"` of this solution.

Use the molar mass of sodium hydroxide to convert the mass of sodium hydroxide to moles.

`90 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(39.997color(red)(cancel(color(black)("g")))) = "2.25 moles NaOH"`

Sodium hydroxide dissociates in a `1:1` mole ratio to produce hydroxide anions, so you can say that you're adding `2.25` moles of hydroxide anions to this solution.

Assuming that the volume of the solution does not change after you dissolve the sample of sodium hydroxide, you can say that the resulting solution will contain `2.25` moles of hydroxide anions in `"1000 mL = 1 L"` of the solution.

This is, of course, equivalent to a molarity of `"2.25 mol L"^(-1)`, which means that you have

`["OH"^(-)] = "2.25 mol L"^(-1)`

Plug this into the equation to find the `"pH"` of the solution

`"pH" = 14 + log(2.25) = color(darkgreen)(ul(color(black)(14.4)))`

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

SIDE NOTE Keep in mind that this is only an approximation of the actual `"pH"` of the solution because calculating the `"pH"` of a solution using the concentration of the hydronium cations--or indirectly by using the concentration of the hydroxide anions--is only accurate for very dilute solutions.

In general, you can use the concentration of the hydronium cations to calculate the `"pH"` of a solution if this concentration does not exceed `"1 mol L"^(-1)`.

The same can be said about the concentration of the hydroxide anions--you can use their concentration to find the `"pOH"`, and thus the `"pH"`, of the solution if their concentration is `< "1 mol L"^(-1)`.

For more concentrated solutions, you need to use the activity of the hydronium cations, not their concentrations*.

`"pH" = - log(a_ ("H"_ 3"O"^(+)))`

`"pOH" = - log(a_("OH"^(-)))`

Here `a_("H"_ 3"O"^(+))` is the activity of the hydronium cations and `a_ ("OH"^(-))` is the activity of the hydroxide anions.