Question

In another experiment, the student titrated 50.0mL of 0.100MHC2H3O2 with 0.100MNaOH(aq) . Calculate the pH of the solution at the equivalence point.

Answer 1

The pH at the equivalence point for the titration of acetic acid with NaOH is determined by the pKa of the acid and the extent of hydrolysis of the conjugate base produced. It is given to be 8.72 according to reference data.

Step-by-step explanation:

The question at hand is concerning the calculation of the pH at the equivalence point during a titration of a weak acid with a strong base, specifically the titration of 50.0 mL of 0.100 M acetic acid (HC₂H₃O₂) with 0.100 M NaOH. To calculate the pH at the equivalence point in such a titration, we should recognize that the weak acid will be completely converted to its conjugate base, which will then undergo hydrolysis with water to form OH- ions.

At the equivalence point, the number of moles of acid will equal the number of moles of base added. This will produce a solution of the conjugate base of the weak acid, which in this case is acetate ions (C₂H₃O₂-). The pH can be determined by using the pKa of acetic acid to find the Kb of the acetate ion and then calculating the pOH from the concentration of OH- produced by the hydrolysis. The pH is then obtained by subtracting the pOH from 14.

From Figure 14.18 (b) provided, the equivalence point for the titration of 25.00 mL of 0.100 M acetic acid with 0.100 M NaOH is given as a pH of 8.72. Hence, for similar concentrations, but double the volume of the acid, we could expect the same pH at the equivalence point.

Answer 2

pH=8.87

Further explanation

Reaction

C₂H₄O₂+NaOH⇒CH₃COONa+H₂O

at the equivalence point = mol C₂H₄O₂= mol NaOH

mol C₂H₄O₂ : 50 x 0.1 = 0.5 mlmol=5.10⁻⁴ mol

The two reactants have completely reacted, and there is only salt(CH₃COONa) and water(H₂O), there will be hydrolysis

For acids from weak acids and strong bases (the solution is alkaline) then the calculation:

`\tt [OH^-]=\sqrt{(Kw)/(Ka)* M }`

M=anion concentration=CH₃COO⁻

Ka=acid constant(for CH₃COOH,Ka=1.8.10⁻⁵)

`\tt [OH^-]=\sqrt{(10^(-14))/(1.8.10^(-5))* 0.1 }`

`\tt [OH^-]=\sqrt{5.6.10^(-11)}=7.483* 10^(-6)`

pOH=6-log 7.483=5.13

pH= 14 - 5.13=8.87