A volume of 100mL of a 0.330M HNO3 solution is titrated with 0.370M KOH. Calculate the volume of KOH required to reach the equivalence point.
HNO3 + KOH ==> KNO3 + H2O
mols HNO3 = M x L = ?
mols KOH = same as mols HNO3 because the ratio of mols is 1:1 from the balanced equation.
Then M KOH = mols KOH/L KOH. You know M and mols, solve for L and change to mL as you wish.
To calculate the volume of potassium hydroxide (KOH) required to reach the equivalence point, we need to use the concept of stoichiometry and the balanced chemical equation of the reaction between HNO3 and KOH.
The balanced equation for the reaction between HNO3 and KOH is as follows:
HNO3 + KOH → KNO3 + H2O
From the balanced equation, we can see that the stoichiometric ratio between HNO3 and KOH is 1:1. This means that for every 1 mole of HNO3, we need 1 mole of KOH to react completely.
Given that the concentration of the HNO3 solution is 0.330 M and the volume used is 100 mL (or 0.100 L), we can calculate the number of moles of HNO3 used using the formula:
moles of HNO3 = concentration × volume
= 0.330 M × 0.100 L
= 0.033 moles
Since HNO3 and KOH react in a 1:1 stoichiometric ratio, we know that the number of moles of KOH required will also be 0.033 moles.
Now, we can calculate the volume of KOH required using its molarity, which is given as 0.370 M:
volume of KOH = moles of KOH / concentration
= 0.033 moles / 0.370 M
= 0.089 L
Converting the volume of KOH to milliliters:
volume of KOH = 0.089 L × 1000 mL/L
= 89 mL
So, the volume of KOH required to reach the equivalence point is 89 mL.