The end point in a titration of a 58mL sample of aqueous HCl was reached by the addition of 25mL of 0.83MNaOH titrant. The reaction proceeds by the following equation. HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l) What is the molar concentration of HCl?

0.36 M

Explanation:

Let's consider the following neutralization reaction.

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

25 mL of 0.83 M NaOH reacted. The moles of NaOH that reacted are:

0.025 L × 0.83 mol/L =0.021 mol

The molar ratio of NaOH to HCl is 1:1. The moles of HCl that reacted are 0.021 moles.

0.021 moles are contained in 58 mL of HCl. The molar concentration of HCl is:

M = 0.021 mol/0.058 L

M = 0.36 M

the answer is c i believe