Goal #2: Measure the pH of a monoprotic acids and its conjugate base Your group

Question

Goal #2: Measure the pH of a monoprotic acids and its conjugate base Your group will be assigned one of the following conjugate acid/base pairs: a. HCl (1.0 M) Naci (58.44 g/mol) b. Formic Acid (1.0 M) Sodium Formate (68.01 g/mol) C Propionic Acid (1.0 M) Sodium propanoate (ge.06gmo d. Ammonium Chloride (53.49 g/mol)-Ammonia (1.0 M) Prepare 0.10 M solutions with volumes of 100 mL of both the assigned acid and its base. For most of the acids (except ammonium) this will involve diluting a 1.0 M solution of the acid (or base in the case of ammonia). For most of the bases (except ammonia) this will involve adding the appropriate mass of the base to a 100 mL volumetric flask and diluting to mark (mix well). Your group will need to determine the appropriate mass for themselves. Measure the pH of the acid and its conjugate base.

Explanation / Answer

To prepare a 1 M solution of propionic acid and 1 M solution of sodium propionate

Propionic acid to be taken = molarity x molar mass x volume of solution

                                           = 1.0 M x 74.1 g/mol x 1 L

                                           = 74.1 g

So dissolving 74.1 g of propionic acid in 1 L of water would give 1 M solution

Sodium propionate to be taken = 1 M x 96.96 g/mol x 1L

                                                  = 96.06 g

So dissolving 96.06 g of sodium propionate in 1 L of water gives 1 M sodium propionate solution

To prepare 0.1 M of 100 ml propionic acid solution

Volume of 1 M Propionic acid solution to be taken = 0.1 M x 100 ml/1.0 M

                                                                                = 10 ml

So mixing 10 ml of 1 M propionic acid with 90 ml water would give 0.1 M propionic acid solution

pH of 0.1 M propionic acid solution

Ka = [H+][propionate]/[propionic acid]

1.34 x 10^-5 = x^2/0.1

x = [H+] = 1.16 x 10^-3 M

pH = -log[H+] = 2.94

To prepare 0.1 M of 100 ml sodium propionate solution

Volume of 1 M sodium propionate solution to be taken = 0.1 M x 100 ml/1.0 M

                                                                                       = 10 ml

So mixing 10 ml of 1 M sodium propionate with 90 ml water would give 0.1 M sodium propionate solution

pH of 0.1 M sodium propanoate solution

Kb = Kw/Ka = [propionic acid][OH-]/[sodium propanoate]

1 x 10^-14/1.34 x 10^-5 = x^2/0.1

x = [OH-] = 8.64 x 10^-6 M

pOH = -log[OH-] = 5.06

pH = 14 - pH = 8.94

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