Question

Question #f9045

Answer 1

You'd neeed 35.3 mL.

Explanation:

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")`