A. How many grams of KOH are needed to neutralize 18.25 g of HCl? b. How many moles of each are involve in this reaction? Thanks
##”28.08 g”##
As you know, potassium hydroxide, ##”KOH”##, is a strong base and hydrochloric acid, ##”HCl”##, is a strong acid, which means that they will both dissociate completely in aqueous solution to form hydroxide anions, ##”OH”^(-)##, and hydrogen cations, ##”H”^(+)##, respectively.
##”KOH”_text((aq]) -> “K”_text((aq])^(+) + “OH”_text((aq])^(-)##
##”HCl”_text((aq]) -> “H”_text((aq])^(+) + “Cl”_text((aq])^(-)##
The hydrogen cations will actually exist as hydronium ions, ##”H”_3″O”^(+)##, in aqueous solution, but you can represent them as ##”H”^(+)## if you want.
These ions will then each other to form water. That is what your reaction is all about – how many hydroxide anions and how many hydrogen cations are present in solution.
The key to this problem is the that exists between these two ions in the balanced chemical equation.
##”KOH”_text((aq]) + “HCl”_text((aq]) -> “KCl”_text((aq]) + “H”_2″O”_text((l])##
Potassium chloride, ##”KCl”##, is a soluble salt, so it will exist as cations and anions in solution. The net ionic equation for this reaction will be
##”OH”_text((aq])^(-) + “H”_text((aq])^(+) -> “H”_2″O”_text((l])##
So, one mole of hydroxide anions will react with one mole of hydrogen cations to form one mole of water.
This tells you that you need equal numbers of moles of strong base and strong acid to reach complete neutralization.
Use hydrochloric acid’s molar mass to determine how many moles you have in that ##”18.25-g”## sample
##18.25 color(red)(cancel(color(black)(“g”))) * “1 mole HCl”/(36.46 color(red)(cancel(color(black)(“g”)))) = “0.5005 moles HCl”##
This means that in order to neutralize the acid, you need to have ##0.5005## moles of strong base.
Use potassium hydroxide’s molar mass to determine how many grams would contain this many moles
##0.5005 color(red)(cancel(color(black)(“moles KOH”))) * “56.106 g”/(1color(red)(cancel(color(black)(“mole KOH”)))) = color(green)(“28.08 g”)##
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