Calculate the pH of the solution that results when 20 mL of 0.2M HCOOH is mixed with 25mL of 0.2M NaOH solution.(post-equivalence point)

The pH of the solution : 12

Reaction

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

mol HCOOH =

mol NaOH =

Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]

ICE method :

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

4                          5

4                          4                     4                   4

0                          1                      1                    1

Concentration of [OH⁻] from NaOH :

pOH=-log[OH⁻]

pOH=-log 10⁻²=2

pH+pOH=14

pH=14-2=12

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.