The pH at the equivalence point for the titration of 0.25 M HCOOH versus 0.25 M NaOH is 8.5.
Let's consider the following neutralization reaction.
HCOOH + NaOH ⇒ HCOONa + H₂O
At the equivalence point, 0.25 M HCOOH completely reacted with 0.25 M NaOH to form 0.25 M HCOONa.
HCOONa undergoes hydrolysis. The net ionic equation is:
HCOO⁻(aq) + H₂O(l) ⇄ HCOOH(aq) + OH⁻(aq)
Given the concentration of HCOO⁻ is 0.25 M (Cb) and the basic dissociation constant of HCOO⁻ is 4.8 × 10⁻¹¹ (Kb), we can calculate the concentration of OH⁻ using the following expression.
[tex][OH^{-} ] = \sqrt{Kb \times Cb } = \sqrt{(4.8 \times 10^{-11} ) \times 0.25 } = 3.5 \times 10^{-6} M[/tex]
The pOH of the solution is:
[tex]pOH = -log [OH^{-} ] = -log (3.5 \times 10^{-6} ) = 5.5[/tex]
The pH of the solution is:
[tex]pH = 14 -pOH = 14 -5.5 = 8.5[/tex]
The pH at the equivalence point for the titration of 0.25 M HCOOH versus 0.25 M NaOH is 8.5.
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