Question

Question #bc8f1

Answer 1

`1 * 10^(-12)` `"M"`

Explanation:

Start by calculating the number of moles of sodium hydroxide present in your sample. To do this, use the molar mass of sodium hydroxide

`0.004 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "0.0001 moles NaOH"`

Now, your first goal here is to calculate the concentration of hydroxide anions, `"OH"^(-)`, in the solution.

You can safely assume that the volume of the solution will be equal to the volume of water, and since sodium hydroxide is a strong base, you can say that your solution will contain `0.0001` moles of hydroxide anions in `"10 mL"` of solution.

This is the case because sodium hydroxide dissociates in a `1:1` mole ratio to produce hydroxide anions, so

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

This means that you will have--remember, by definition, the molarity of the hydroxide anions must reflect a volume of `10^3color(white)(.)"mL" = "1 L"` of solution!

`["OH"^(-)] = "0.0001 moles NaOH"/(10 * 10^(-3)color(white)(.)"L") = "0.01 mol L"^(-1)`

Now, an aqueous solution at room temperature has--I won't add the units for `K_W = 10^(-14)`

`["OH"^(-)] * ["H"_ 3"O"^(+)] = K_W = 10^(-14)`

Here `["H"_3"O"^(+)]` is the concentration of hydronium cations, which you'll sometimes see written as hydrogen cations, `"H"^(+)`, and `K_W` is the ionization constant of water.

This means that you have

`["H"_3"O"^(+)] = 10^(-14)/(["OH"^(-)])`

Plug in your value to find

`["H"_3"O"^(+)] = 10^(-14)/(0.01) = color(darkgreen)(ul(color(black)(1 * 10^(-12)color(white)(.)"M")))`

The answer is rounded to one significant figure.