The molarity of an aqueous solution of potassium hydroxide, KOH, is determined by titration against a 0.197 M HI solution. If 28.4 mL of the base are required to neutralize 18.6 mL of HI, what is the molarity of the KOH solution?
Assuming HI + KOH = KI + H2O,
one mole of KOH is needed to neutralize ach mole of HI
18.6ml of .197M HI contains .00366 moles of HI
So, we need .00366 moles of KOH in 28.4mL of solution.
.0284x = .00366
x = .129, so the KOH is .129M
To determine the molarity of the KOH solution, we need to use the concept of stoichiometry and the balanced equation for the reaction between KOH and HI.
The balanced equation for the reaction is:
KOH + HI -> KI + H2O
From the balanced equation, we can see that the mole ratio between KOH and HI is 1:1. This means that 1 mole of KOH reacts with 1 mole of HI.
First, let's calculate the number of moles of HI used in the titration:
moles of HI = Molarity × Volume
= 0.197 M × 0.0186 L
= 0.0036542 moles
Since the mole ratio is 1:1, the number of moles of KOH used in the titration is also 0.0036542 moles.
Now, let's calculate the molarity of the KOH solution:
Molarity of KOH = moles of KOH / Volume of KOH solution
= 0.0036542 moles / 0.0284 L
≈ 0.1284 M
Therefore, the molarity of the KOH solution is approximately 0.1284 M.