To determine the amount of sodium hydroxide (NaOH) and acetic acid needed to find the solubility of sodium acetate, you need to use stoichiometry and the balanced chemical equation for the reaction between these compounds.
The balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) to form sodium acetate (CH3COONa) and water (H2O) is:
CH3COOH + NaOH → CH3COONa + H2O
From the balanced equation, you can see that the stoichiometric ratio between acetic acid and sodium hydroxide is 1:1. This means that for every 1 mole of NaOH, you need 1 mole of acetic acid to react completely.
In your case, you have 0.100 mol of NaOH, so you would also need 0.100 mol of acetic acid to react completely with it. However, you have 0.837 mol of acetic acid, which is in excess.
To determine the amount of sodium acetate that will precipitate, you need to calculate the limiting reactant. In this case, the limiting reactant is sodium hydroxide because it is present in a smaller amount.
To find the solubility of sodium acetate, you will need to heat up the solution of acetic acid and sodium hydroxide mixture until a precipitate forms. The precipitate will be sodium acetate, which can then be weighed to determine its mass.
To calculate the mass of sodium acetate that will precipitate, you first need to find the number of moles of sodium acetate formed. Since the stoichiometric ratio between sodium hydroxide and sodium acetate is also 1:1, the number of moles of sodium acetate will be the same as the number of moles of sodium hydroxide used.
So, the mass of sodium acetate that will precipitate can be calculated using the formula:
mass (sodium acetate) = moles (sodium hydroxide) x molar mass (sodium acetate)
Finally, to find the solubility of sodium acetate, you can divide the mass of sodium acetate that will precipitate by the volume of the solution.
solubility = mass (sodium acetate) / volume (solution)
Remember to convert the volume of the solution to the same units as the mass of sodium acetate.
Keep in mind that this explanation assumes ideal conditions and complete reaction. In practice, the solubility might be slightly different due to factors such as temperature, pressure, impurities, and other variables.