Question

To neutralize a 20.0 ml sample of 0.750 m hcl using titration, what volume (ml) of the 0.750 m naoh would you need at the equivalence point of the titration?

Answer 1

To neutralize a 20.0 mL sample of 0.750 M HCl at the equivalence point of a titration, 20.0 mL of 0.750 M NaOH is required, due to the 1:1 molar ratio in the neutralization reaction.

Step-by-step explanation:

To determine the volume of 0.750 M NaOH needed to neutralize a 20.0 mL sample of 0.750 M HCl at the equivalence point of the titration, we can use the stoichiometry of the neutralization reaction. The reaction between HCl and NaOH is given by the balanced chemical equation:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

According to the balanced equation, 1 mole of HCl reacts with 1 mole of NaOH. In this titration, both HCl and NaOH have the same molarity (0.750 M). Since the volume of the HCl solution is 20.0 mL, we use the concept that at equivalence point, the moles of acid will equal the moles of base to neutralize each other completely.

Therefore, to neutralize 20.0 mL of 0.750 M HCl, we will also need 20.0 mL of 0.750 M NaOH, as the stoichiometric coefficients of the acid and base in the reaction are equal (1:1 molar ratio).