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

Answer:

Explanation:

moles of acetic acid = 500 x 10⁻³ x .1 M

= 5 X 10⁻³ M

.005 M

Moles of NaOH = .1 M

Moles of sodium acetate formed = .005 M

Moles of NaOH left = .095 M

pOH = 4.8 + log .005 / .095

= 4.8 -1.27875

= 3.52125

pH = 14 - 3.52125

= 10.48