A25.0 ml sample of aqueous sodium hydroxide has been used to titrate to the second equivalence point 22.30 ml of 0.253 m sulfuric acid. what is the molarity of the sodium hydroxide solution?

The molarity of the sodium hydroxide is calculated by first calculating the number of moles of sulfuric acid in the 22.30 ml used.

This is done as follows since 0.253M is contained in 1000ml solution.

(0.253×22.3)/1000=0.00564 moles

The equation for the reaction is:

H₂SO₄₍ₐq)+2NaOH₍ₐq)⇒ Na₂SO₄₍aq) +2 H₂O₍l₎

Therefore the reacting ratios between sodium hydroxide and sulfuric acid is 2:1

therefore the number of moles that reacted with the sulfuric acid is calculated as follows:

(0.00564 moles ₓ 2)/1=0.01128moles

0.1128moles is in 25ml therefore, 1000ml has:

(1000ₓ0.01128moles)/25= 0.4512M NaOH

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.