this is probably a simple proportion but i’m having trouble with it
find the number of mL of 0.0144 M HCl needed to completely neutralize 420mg of KOH
✅ Answers
? Best Answer
The equation for this neutralization is:
KOH + HCl (aq) –> KCl + H2O
One mole of potassium hydroxide reacts with one mole of HCl to yield a salt potassium chloride and water.
The molecular weight of KOH = 56.11 grams / mole
420 mg of KOH = 0.420 grams
0.420 grams / 56.11 grams per mole = 0.7485 moles
So you need 0.7485 moles of HCl to neutralize.
Each 10 mL of 0.0144 M HCl gives you 0.0144 moles of HCl.
(0.7485 x 10) / 0.0144 = 520 mL