Question

Sodium benzoate, the sodium salt of benzoic acid, is used as a food preservative. A sample containing solid sodium benzoate mixed with sodium chloride is dissolved in 50.0 mL of 0.500 M HCl, giving an acidic solution (ben- zoic acid mixed with HCl). This mixture is then titrated with 0.393 M NaOH. After the addition of 46.50 mL of the NaOH solution, the pH is found to be 8.2. At this point, the addition of one more drop (0.02 mL) of NaOH raises the pH to 9.3. Calculate the mass of sodium benzo- ate (NaC6H5COO) in the original sample. (Hint: At the equivalence point, the total number of moles of acid [here HCl] equals the total number of moles of base [here, both NaOH and NaC6H5COO].)

Answer 1

0.966g of sodium benzoate are in the original sample

Step-by-step explanation:

The sodium benzoate, B⁻, reacts with HCl as follows:

B⁻ + HCl → BH + Cl⁻

Where 1 mole of benzoate reacts per mole of HCl

The HCl is added in excess. We can find the HCl that reacts with sodium benzoate using the NaOH that reacts with HCl.

When 1 drop change the pH in 1,1 units, this point is the equivalence point. Thus, moles in excess of HCl = Moles of NaOH added are:

0.04650L * (0.393mol / L) = 0.0183 moles NaOH = Moles HCl in excess

Moles HCl added:

0.0500L * (0.500mol/L) = 0.0250 moles HCl added

Moles HCl that react = Moles Sodium benzoate:

0.0250 moles HCl - 0.0183 moles HCl = 0.0067 moles HCl = Moles Sodium benzoate

Mass Sodium benzoate -Molar mass: 144.11g/mol-:

0.0067 moles Sodium benzoate * (144.11g/mol) =

0.966g of sodium benzoate are in the original sample