# Question f9045

Jul 24, 2015

You'd neeed 35.3 mL.

#### Explanation:

Start by writing the balanced chemical equation for this neutralization reaction

$H {I}_{\left(a q\right)} + K O {H}_{\left(a q\right)} \to K {I}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

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 $K O H$ took part in the reaction

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{K O H} = \text{0.171 M" * 27.9 * 10^(-3)"L" = "0.004771 moles}$ $K O H$

This means that you also had

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

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

$C = \frac{n}{V} \implies V = \frac{n}{C}$

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

Expressed in mililiters, the answer will be

${V}_{H I} = \textcolor{g r e e n}{\text{35.3 mL}}$