Question #bc8f1

1 Answer
Aug 11, 2017

Answer:

#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.