Question #f7407
1 Answer
Explanation:
The first thing that you need to do here is to calculate the molarity of the hydroxide anions,
#"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
#["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"#