For each of the following primary standard acids, write the titration equation and
calculate the volume of 0.100 M NaOH required for the titration:
(i) 300 mg benzoic acid (C6H5COOH) (ii) 200 mg oxalic acid dihydrate
(C2O4H2.2H2O)
Equations: (i): C6H5COOH +NaOH->NaC6H5COO +H2
(ii): H2C2O4.2H2)+2NaOH->>Na2C2O4+4H2O
To calculate the volume of NaOH required for the titration, we need to use the concept of stoichiometry. The stoichiometry of a balanced chemical equation tells us how many moles of reactants are required to react completely.
For both equations, the stoichiometric ratio between benzoic acid and NaOH is 1:1, and between oxalic acid dihydrate and NaOH, it is 2:2. This means that for every 1 mole of benzoic acid or oxalic acid dihydrate, we need 1 mole or 2 moles of NaOH, respectively.
First, let's calculate the moles of benzoic acid in 300 mg (0.300 g). The molar mass of benzoic acid (C6H5COOH) is:
(6 * 12.01 g/mol) + (5 * 1.01 g/mol) + 12.01 g/mol + 16.00 g/mol + 16.00 g/mol = 122.12 g/mol
Now, we can calculate the moles of benzoic acid:
moles = mass / molar mass = 0.300 g / 122.12 g/mol
Next, let's calculate the volume of 0.100 M NaOH required. To do this, we need to use the equation:
moles = volume (L) * concentration (mol/L)
We can rearrange this equation to solve for volume:
volume (L) = moles / concentration (mol/L)
Substituting the values:
volume (L) = moles / 0.100 mol/L
Now, let's calculate the moles of NaOH required:
moles of NaOH = moles of benzoic acid (since the stoichiometric ratio is 1:1)
Finally, we can calculate the volume of NaOH required:
volume (L) = moles of NaOH / 0.100 mol/L
Follow the same steps to calculate the volume of NaOH required for oxalic acid dihydrate.