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To a 25.00-mL volumetric flask, a lab technician adds a 0.100 g sample of a weak monoprotic acid, HA, and dilutes to the mark with distilled water. The technician then titrates this weak acid solution with 0.0903 M KOH. She reaches the endpoint after adding 47.51 mL of the KOH solution. Determine the number of moles of the weak acid in the solution. Determine the molar mass of the weak acid. After the technician adds 15.99 mL of the KOH solution, the pH of the mixture is 5.29. Determine the pKa of the weak acid.

Respuesta :

The pH of the mixture is 5.29. Determine the pKa of the weak acid is 5.58

A weak acid is an acid that partially dissociates into its ions in an aqueous solution or water. In assessment, a robust acid absolutely dissociates into its ions in water. The conjugate base of a weak acid is a vulnerable base, at the same time as the conjugate acid of a weak base is a susceptible acid.

47.51 ml of 0.0903 M KOH = 0.04751 L × 0.0903 mol/L

                                            = 4.290 × 10 ⁻³ mole

∴ A number of the of weak acids = 4.29  × 10 ⁻³ mole

as one mole of KOH neutralizes the one-mole weak acid

∴ The molar mass of weak acid = mass/mole

                                             = 0.100 g / 4.29 × 10⁻³

                                             = 23.31 g/mol

15.99 ml of 0.0903 M KOH = 0.01599L × 0.0903 mol/L

                                            = 1.44 × 10⁻³

Thus after addition of KOH

mole of salt = 1.44 × 10⁻³

A mole of acid = 4.29  × 10⁻³ -  1.44 × 10⁻³

mole of acid = 2.85 × 10⁻³

Henderson equation

pH = PKa + log(salt)/(acid)

5.29 = PKa + log (1.44 × 10⁻³)/(2.85 × 10⁻³)

5.29 = PKa + (-0.296)

PKa = 5.58

Learn more about weak acid here:-https://brainly.com/question/24018697

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