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

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.