Answer:
0.966g of sodium benzoate are in the original sample
Explanation:
The sodium benzoate, B⁻, reacts with HCl as follows:
B⁻ + HCl → BH + Cl⁻
<em>Where 1 mole of benzoate reacts per mole of HCl</em>
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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
<em>Moles HCl added:</em>
0.0500L * (0.500mol/L) = 0.0250 moles HCl added
<em>Moles HCl that react = Moles Sodium benzoate:</em>
0.0250 moles HCl - 0.0183 moles HCl = 0.0067 moles HCl = Moles Sodium benzoate
<em>Mass Sodium benzoate -Molar mass: 144.11g/mol-:</em>
0.0067 moles Sodium benzoate * (144.11g/mol) =
<h3>0.966g of sodium benzoate are in the original sample</h3>