Question #f7407

1 Answer
Stefan V.
Jan 14, 2018

#["H"_3"O"^(+)] = 1 * 10^(-14)# #"M"#

Explanation:

The first thing that you need to do here is to calculate the molarity of the hydroxide anions, #"OH"^(-)#, which are delivered to the solution by the sodium hydroxide in a #1:1# mole ratio.

#"moles of OH"^(-) = "moles of NaOH"#

To do that, use the molar mass of sodium hydroxide to convert the number of grams of solute to moles

#4 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "0.1 moles NaOH"#

This means that your solution will contain #0.1# moles of hydroxide anions, which in #"100 mL" = 100/1000 quad "L"# of the solution have a molarity of

#["OH"^(-)] = "0.1 moles OH"^(-)/(100 * 10^(-3) quad "L")= "1 M"#

Now, you know that an aqueous solution at room temperature has

#color(blue)(ul(color(black)(["H"_3"O"^(+)] * ["OH"^(-)] = 1 * 10^(-14)quad "M"^2)))#

This means that your solution contains

#["H"_3"O"^(+)] = (1 * 10^(-14)"M"^color(red)(cancel(color(black)(2))))/(1color(red)(cancel(color(black)("M")))) = 1 * 10^(-14)quad "M"#

Related questions
See all questions in pH
Impact of this question
1350 views around the world
You can reuse this answer
Creative Commons License