Question #f9045

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Answer 1 · 2015-07-24T08:36:03

You'd neeed 35.3 mL.

Start by writing the balanced chemical equation for this neutralization reaction

$HI_((aq)) + KOH_((aq)) -> KI_((aq)) + H_2O_((l))$

The important thing to notice here is that you ahve a $1:1$ mole ratio between hydroiodic acid and potassium hydroxide.

This means that, in order to have a complete neutralization, you need equal numbers of moles of each compound.

SInce you know the molarity and volume of the potassium hydroxide solution, you can determine how many moles of $KOH$ took part in the reaction

$C = n/V => n = C * V$

$n_(KOH) = "0.171 M" * 27.9 * 10^(-3)"L" = "0.004771 moles"$ $KOH$

This means that you also had

$0.004771cancel("moles"KOH) * ("1 mole "HI)/(1cancel("mole"KOH)) = "0.004771 moles"$ $HI$

Now simply use the hydroiodic acid solution's molarity to see what volume would contain this many moles

$C = n/V => V = n/C$

$V_(HI) = (0.004771cancel("moles"))/(0.135cancel("moles")/"L") = "0.0353 L"$

Expressed in mililiters, the answer will be

$V_(HI) = color(green)("35.3 mL")$