This question is based on an experiment to determine the mass of ASA.
Information from the experiment :
• ASA is a weak acid with a Ka value of 3.2 x 10^-4.
• An ASA tablet has an approximate mass of 500 mg.
• The experiment used a titration with a standardized 0.100 mol/L NaOH.
What volume of base will you substitute into the formula when you calculate the number of moles of base used? I know from looking at the balanced equation that NaOH reacts 1:1 molar with the ASA. HC9H7O4(aq) + NaOH(aq) NaC9H7O4(aq) + H2O(l)
I am just unsure of how to calculate what volume of base to substitute. Any help is appreciated.
To determine the volume of base (NaOH) to substitute into the formula when calculating the number of moles of base used in the titration, you will need to use the concept of stoichiometry and the equation balanced equation provided.
First, let's understand the stoichiometry of the reaction based on the balanced equation:
HC9H7O4(aq) + NaOH(aq) -> NaC9H7O4(aq) + H2O(l)
The balanced equation shows that one mole of ASA (HC9H7O4) reacts with one mole of NaOH to produce one mole of NaC9H7O4 and one mole of water.
Now, let's use the given information to calculate the volume of NaOH used in the experiment:
1. You know that the concentration of NaOH used in the experiment is 0.100 mol/L. This means that in every liter of the NaOH solution, there are 0.100 moles of NaOH.
2. You also know the molar ratio of ASA to NaOH is 1:1. This means that for every mole of ASA reacted, you need one mole of NaOH.
3. The experiment used an ASA tablet with an approximate mass of 500 mg. To calculate the number of moles of ASA, divide the mass by its molar mass. The molar mass of ASA is calculated based on its constituent elements: Hydrogen (H) = 1.01 g/mol, Carbon (C) = 12.01 g/mol, and Oxygen (O) = 16.00 g/mol. Since ASA has 9 carbons, 8 hydrogens, and 4 oxygens, the molar mass of ASA is (9 * 12.01) + (8 * 1.01) + (4 * 16.00) = 180.16 g/mol. Therefore, the number of moles of ASA is (500 mg / 1000 mg/g) / 180.16 g/mol = 0.00277 moles.
4. Using the molar ratio from the balanced equation (1:1), the number of moles of NaOH is also 0.00277 moles.
5. Finally, to determine the volume of NaOH used, use the equation:
Volume (L) = Moles / Concentration (mol/L)
Volume = 0.00277 moles / 0.100 mol/L
Volume = 0.0277 L or 27.7 mL
Therefore, the volume of NaOH used in the experiment is approximately 27.7 mL, which should be substituted into the formula when calculating the number of moles of base used.
Are you titrating the ASA with NaOH? If so use the volume NaOH in the titration.
Then mols = M NaOH x L NaOH