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##”0.207 g KOH”##
Your strategy here will be to
So, you know that a ##”0.118 M”## solution of hydrochloric acid will contain ##0.118## moles of acid per liter of solution. This means that your
##31.32 color(red)(cancel(color(black)(“mL”))) * “1 L”/(10^3color(red)(cancel(color(black)(“mL”)))) = “0.03132 L”##
sample will contain
##0.03132 color(red)(cancel(color(black)(“L solution”))) * “0.118 moles HCl”/(1color(red)(cancel(color(black)(“L solution”)))) = “0.003696 moles HCl”##
Potassium hydroxide and hydrochloric acid react in a ##1:1## mole ratio to produce aqueous potassium chloride and water
##”KOH”_ ((aq)) + “HCl”_ ((aq)) -> “KCl”_ ((aq)) + “H”_ 2″O”_ ((l))##
This means that a complete requires equal numbers of moles of acid and of base. You can thus say that the sample of potassium hydroxide must have contained ##0.003696## moles of potassium hydroxide, since that’s how many moles of hydrochloric acid were consumed in the reaction.
Finally, potassium hydroxide has a molar mass of ##”56.11 g mol”^(-1)##, i.e. one mole of this compound has a mass of ##”56.11 g”##.
This means that the mass of potassium hydroxide dissolved in solution was
##0.003696 color(red)(cancel(color(black)(“moles KOH”))) * “56.11 g”/(1color(red)(cancel(color(black)(“mole KOH”)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(“0.207 g”)color(white)(a/a)|)))##
The answer is rounded to three , the number of sig figs you have for the of the hydrochloric acid solution.
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