contestada

) 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.

Respuesta :

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

mickymike92

The pH at equivalence point is 8.87

Equation of the reaction

C₂H₄O₂ + NaOH ⇒ CH₃COONa + H₂O

What happens at equivalence point?

At the equivalence point, the moles of acid equals the moles of base, so the acid and base are completely neutralized.

moles of C₂H₄O₂ = moles of NaOH

  • moles = molarity × volume

volume of C₂H₄O₂ = 50.0 mL = 0.05 L

moles of C₂H₄O₂ = 0.1 × 0.05

moles of C₂H₄O₂ = 0.005 moles

At equivalence point, only salt (CH₃COONa) and water(H₂O), are present and salt hydrolysis occurs.

For salts made from weak acids and strong bases, the solution produced is alkaline.

Calculating the hydroxide ion concentration, [OH-]

The formula for determining [OH-] is given as:

[OH-] = √(Kw/Ka × M)

where:

M is anion concentration (CH₃COO⁻) = 0.1 M

Ka for CH₃COOH = 1.8.10⁻⁵

Therefore,

[OH-] = √(10^-14/1.8.10⁻⁵ × 0.1)

[OH-] = 7.483 × 10^-6

How can pH be calculated from [OH-]?

The formula relating pH and pOH is given below:

  • pOH + pH = 14

pOH = -log[OH-]

pOH = -log 7.483 × 10^-6

pOH = 5.13

Hence; from pOH + pH = 14

pH = 14 - pOH

pH = 14 - 5.13

pH = 8.87

Therefore, the pH at equivalence point is 8.87

Learn more about equivalence point and pH at: https://brainly.com/question/25487920

Otras preguntas