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

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:

$\tt [OH^-]=\sqrt{\dfrac{Kw}{Ka}\times M }$

M=anion concentration=CH₃COO⁻

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

$\tt [OH^-]=\sqrt{\dfrac{10^{-14}}{1.8.10^{-5}}\times 0.1 }$

$\tt [OH^-]=\sqrt{5.6.10^{-11}}=7.483\times 10^{-6}$

pOH=6-log 7.483=5.13

pH= 14 - 5.13=8.87