Given that delta Hrxn for OH- ions is -22.9 kJ/mol, calculate the enthalpy of neutralization when I mole of a strong monoprotic acid (such as HCl) is titrated by 1 mole of a strong base (such as KOH) at 25-degrees C.
To calculate the enthalpy of neutralization, we need to use the balanced chemical equation for the reaction between the acid (HCl) and the base (KOH). The balanced equation is:
HCl (aq) + KOH (aq) → H2O (l) + KCl (aq)
We know that for 1 mole of OH- ions, the enthalpy change (ΔHrxn) is -22.9 kJ/mol. Since KOH is a strong base and it dissociates completely in water, we can assume that 1 mole of KOH will yield 1 mole of OH- ions.
In the reaction, 1 mole of HCl is reacting with 1 mole of KOH. Therefore, the enthalpy change for this reaction can be calculated as follows:
ΔHneutralization = ΔHrxn × n
where ΔHneutralization is the enthalpy of neutralization, ΔHrxn is the enthalpy change per mole of OH- ions (-22.9 kJ/mol), and n is the number of moles of OH- ions involved in the reaction.
In this case, n = 1 mole of OH- ions.
Substituting the values into the equation:
ΔHneutralization = -22.9 kJ/mol × 1 mol = -22.9 kJ
Therefore, the enthalpy of neutralization when 1 mole of a strong monoprotic acid (HCl) is titrated by 1 mole of a strong base (KOH) at 25 degrees C is -22.9 kJ.