Using a BCA table, calculate the molarity of a 25.0 mL sample of HNO3 that's is titrated to its endpoint with 22.7 mL of 0.200 M NaOH.
HNO3 + NaOH ==> NaNO3 + HOH
You construct the table.
mols NaOH = M x L = ??
The equation tells you it is 1 mol NaOH to 1 mol HNO3; therefore.
mols NaOH = mols HNO3
Then M HNO3 = mols HNO3/L HNO3.
Yo have mols and you have L. Calculate M HNO3.
thanks
To calculate the molarity of the HNO3 solution, we need to use the balanced chemical equation, known volumes of the HNO3 solution and NaOH solution, and their respective molarities.
The balanced chemical equation between HNO3 (Nitric acid) and NaOH (Sodium hydroxide) is as follows:
HNO3 + NaOH -> NaNO3 + H2O
From the equation, we can see that the stoichiometric ratio between HNO3 and NaOH is 1:1. This means that 1 mole of HNO3 reacts with 1 mole of NaOH.
To use the BCA (Before, Change, After) table:
1. Write down the initial concentration (Before) and volume for each species:
- Initial volume of HNO3 solution (Before): 25.0 mL = 0.025 L (converted to liters)
- Initial concentration of NaOH solution (Before): 0.200 M
- Initial volume of NaOH solution (Before): 22.7 mL = 0.0227 L (converted to liters)
2. Determine the species that will limit the reaction. In this case, it is NaOH since it is the limiting reagent since it reacts with HNO3 in a 1:1 ratio.
3. Use the stoichiometric ratio between NaOH and HNO3 to determine the amount (Change) of HNO3 reacted:
- NaOH: 0.200 M x 0.0227 L = 0.00454 mol
Since the stoichiometric ratio is 1:1, the amount (Change) of HNO3 reacted is also 0.00454 mol.
4. Calculate the final concentration (After) of HNO3 by dividing the amount of HNO3 reacted by the final volume:
- Final volume of HNO3 solution (After): 25.0 mL + 22.7 mL = 47.7 mL = 0.0477 L (converted to liters)
- Final concentration of HNO3 (After): 0.00454 mol / 0.0477 L = 0.095 M
Therefore, the molarity of the 25.0 mL sample of HNO3 solution is approximately 0.095 M.