If 43.2 mL of 2.46 M HClO4 solution completely reacts with a 75.0 mL sample of a sodium hydroxide solution, what is the molarity of the sodium hydroxide solution? (hint: write the complete balanced rxn)
HClO4 + NaOH ==> NaClO4 + H2O
mols HClO4 = M x L = ?
Using the coefficients in the balanced equation (they are 1:1), convert mols HClO4 to mols NaOH.
Now M NaOH = mols/L. You know mols and L, solve for M
To find the molarity of the sodium hydroxide (NaOH) solution, we need to use the balanced chemical equation for the reaction between HClO4 and NaOH.
The balanced equation for the reaction between HClO4 and NaOH is:
HClO4 + NaOH -> NaClO4 + H2O
From the balanced equation, we can see that the stoichiometric ratio between HClO4 and NaOH is 1:1. This means that 1 mole of HClO4 reacts with 1 mole of NaOH.
To find the moles of HClO4, we can use the formula:
moles = concentration x volume (in liters)
Given that the volume of the HClO4 solution is 43.2 mL and the concentration is 2.46 M, we can calculate the moles of HClO4 as follows:
moles of HClO4 = 2.46 M x 0.0432 L = 0.106272 moles
Since the stoichiometric ratio is 1:1, the moles of NaOH in the reaction are also 0.106272 moles.
Now, we can find the molarity of the NaOH solution by dividing the moles of NaOH by the volume of the NaOH solution in liters.
Given that the volume of the NaOH solution is 75.0 mL, we need to convert it to liters by dividing by 1000:
volume of NaOH solution = 75.0 mL / 1000 = 0.075 L
Now we can calculate the molarity of the NaOH solution:
molarity of NaOH = moles of NaOH / volume of NaOH solution
molarity of NaOH = 0.106272 moles / 0.075 L
By performing this calculation, we find that the molarity of the sodium hydroxide solution is 1.417 M (rounded to three decimal places).