27.8 mL solution of 0.797 M HCHO2 with 0.928 M NaOH. What is the pH for the solution at the equivalence point in the titration?

8.69 is the pH at the equivalence point

Explanation:

Formic acid, HCHO₂, reacts with NaOH as follows:

HCHO₂ + NaOH → NaCHO₂ + H₂O.

At the equivalence point you will have in the reaction just NaCHO₂ and H₂O. The concentration of NaCHO₂ will be:

Moles: 0.0278L * 0.797mol/L = 0.02216moles

To reach the equivalence point it is necessary to add:

0.02216mol * (1L / 0.928mol) = 0.0239L

Total volume in the equivalence point:

0.0278L + 0.0239L = 0.0517L

Concentration: 0.02216moles / 0.0517L = 0.429M

The equilibrium of NaCHO₂, CHO₂⁻, in water is:

CHO₂⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + HCHO₂(aq)

Where Kb, 5.56x10⁻¹¹ is defined as:

5.56x10⁻¹¹ = [OH⁻] [HCHO₂] / [CHO₂⁻]

In the equilibrium, it is produced X OH⁻ and HCHO₂, and as concentration of NaCHO₂ is 0.429M:

5.56x10⁻¹¹ = [X] [X] / [0.429M]

2.383x10⁻¹¹ = X²

4.88x10⁻⁶ = X = [OH⁻]

As pOH = -log [OH⁻]

pOH = 5.31

And pH = 14 - pH

pH = 8.69 is the pH at the equivalence point

there all right good

explanation: