What is the molarity of an aqueous solution of potassium hydroxide if 21.34 mL is exactly neutralized by 20.78 mL of 0.116 M HCl by the following reaction?
2NaOH(aq) + H2SO4(aq)-->Na2SO4 2H2O
I simply want to know how to work the problem.
I already have the answer anyways.
Nvm, I got it.
Anonymous
Favorite Answer
I know it but isn't that cheating???
rubidium.chloride
This is how to do it. Here's the corrected equation (your equation is wrong, though).
KOH + HCl --> KCl + H2O
First, convert the volumes to L. Dividing by 1000, we have 0.02134 L and 0.02078 L respectively.
Determine the number of moles present in 0.02078 L of HCl. That way, we can later on convert to moles of KOH using the mole ratio between HCl and KOH.
0.02078 L HCl * 0.116 mol HCl / 1 L
Multiply by the mole ratio between HCl and KOH. The mole ratio is 1:1. You can get these ratios from the coefficients of the balanced equation. Remember to put the unknown (the unit you want to get) on top and the known at the bottom (the given).
0.02078 L HCl * 0.116 mol HCl / 1 L * 1 mol KOH / 1 mol HCl = 0.00241 mol KOH
After getting the number of moles KOH, divide this by the volume of KOH (0.02134 L). This will give you the molarity of KOH
M = 0.00241 mol / 0.02134
= 0.113 mol
:)
Ahmed A
I know but thats cheating with your family/