In the titration of 45 mL of liquid drain cleaner containing NaOH, 25 mL of 0.75 M HNO3 must be added to reach the equivalence point. What is the molarity of the base in the cleaner?
NaOH + HNO3 ==> NaNO3 + H2O
mols HNO3 = M x L = 0.75 M x 0.045 L = ?
From the equation you know 1 mol NaOH = 1 mol HNO3; therefore,
mols NaOH = mols HNO3 from above.
M NaOH = mols NaOH/L NaOH = ?
Post your work if you get stuck.
To determine the molarity of the base in the liquid drain cleaner, we can use the concept of stoichiometry and the volume and molarity of the acid used in the titration.
First, let's write the balanced chemical equation for the reaction between sodium hydroxide (NaOH) and nitric acid (HNO3):
NaOH + HNO3 → NaNO3 + H2O
From the balanced equation, we can see that the stoichiometry between NaOH and HNO3 is 1:1. This means that one mole of NaOH reacts with one mole of HNO3.
Given that 25 mL of 0.75 M HNO3 is used, we can calculate the number of moles of HNO3 used:
moles of HNO3 = volume (L) × molarity (mol/L)
= (25 mL / 1000 mL/L) × 0.75 mol/L
= 0.01875 mol
Since the stoichiometry between NaOH and HNO3 is 1:1, the number of moles of NaOH in the reaction is also 0.01875 mol.
Now, let's calculate the molarity of the base (NaOH) in the liquid drain cleaner:
molarity of NaOH = moles of NaOH / volume of NaOH (L)
= 0.01875 mol / (45 mL / 1000 mL/L)
= 0.4167 M
Therefore, the molarity of the base (NaOH) in the liquid drain cleaner is approximately 0.4167 M.
To determine the molarity of the base in the drain cleaner, we can use the concept of stoichiometry and the equation of the reaction between NaOH and HNO3.
The balanced equation for the reaction is:
NaOH + HNO3 → NaNO3 + H2O
From the equation, we can see that the mole ratio between NaOH and HNO3 is 1:1. This means that for every mole of NaOH, we need one mole of HNO3 to reach the equivalence point.
First, let's calculate the number of moles of HNO3 used in the titration:
moles of HNO3 = volume of HNO3 (L) × molarity of HNO3 (mol/L)
= 0.025 L × 0.75 mol/L
= 0.01875 mol
Since the mole ratio between NaOH and HNO3 is 1:1, this also represents the number of moles of NaOH in the drain cleaner.
Next, let's calculate the molarity of the NaOH in the drain cleaner:
molarity of NaOH = moles of NaOH / volume of NaOH (L)
= moles of NaOH / 0.045 L
= 0.01875 mol / 0.045 L
≈ 0.4167 mol/L
Therefore, the molarity of the base (NaOH) in the drain cleaner is approximately 0.4167 mol/L.