A 20.00 mL sample of a KOH solution required 31.32 mL of 0.118 M HCl for neutralization. What mass (g) of KOH was present in the 20.00 mL of the sample of KOH - (Molar mass of KOH 56.11 g/mol)?

Your strategy here will be to

• calculate the number of moles of hydrochloric acid, #"HCl#, that were consumed in the reaction
• use the mole ratio that exists between the two reactants to calculate the number of moles of potassium hydroxide, #"KOH"#, present in the sample
• use the molar mass of potassium hydroxide to convert the number of moles to grams

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 neutralization 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 sig figs, the number of sig figs you have for the molarity of the hydrochloric acid solution.