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QUESTION

# 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)?

"0.207 g KOH"

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 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.