What is the molarity (concentration) of 2.53 L of NaOH which exactly neutralizes 5.0L of 1.25 M HCl?

The concentration of NaOH is 2.47 M

Explanation:

The equation of the reaction is

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

Given;

Concentration of acid CA= 1.25 M

Volume of acid VA= 5.0L

Concentration of base CB=

Volume of base VB= 2.53 L

Number of moles of acid NA= 1

Number of moles of base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB= CBVBNA

CB= CAVANB/VBNA

CB= 1.25 × 5.0 ×1 / 2.53×1

CB= 2.47 M

Therefore the concentration of NaOH is 2.47 M

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.