Question

1.​Write reaction equations to explain how your acetic acid-acetate buffer reacts with an acid and reacts with a base. 2.​Buffer capacity has a rather loose definition, yet it is an important property of buffers. A commonly seen definition of buffer capacity is: “The amount of H+ or OH– that can be neutralized before the pH changes to a significant degree.” Use your data to determine the buffer capacity of Buffer A and Buffer B. 3.​Say, for example, that you had prepared a Buffer C, in which you mixed 8.203 g of sodium acetate, NaC2H3O2, with 100.0 mL of 1.0 M acetic acid. a. What would be the initial pH of Buffer C? b. If you add 5.0 mL of 0.5 M NaOH solution to 20.0 mL each of Buffer B and Buffer C, which buffer’s pH would change less? Explain.

Answer 1

Q1.

acetic acid = CH3COOH(aq) , acetate = CH3COO-(aq) and proton = H+(aq)

so...

CH3COOH(aq) <-> CH3COO-(aq) + H+(aq)

this is the dissociation of acid, therefore, it will yield H+ ions in solution, which are ACIDIC

then... for the acetate buffer, recall that

CH3COO-(aq) is a weak conjugate base, so, it can accept H+ if in excess

so

H+(aq) + CH3COO-(aq) <--> CH3COOH(aq)

therefore, addition of excess H+, will form CH3COOH slowly.. i.e. th epH will not increase drstically

Addition of OH- ins, from a strong base:

OH-(aq) + H+(aq) <--> H2O(l)

then, OH- is apdily converted to H2O, which is not basic, so pH won't change drastically.

Q2.

Please state Buffer A + B compositions

Q3.

m = 8.203 g of NaCH3COO

V = 100 mL of CH3COOH

a=

initial pH of buffer C:

apply buffer eqution from Henderson Hasselbach

pH = pKa + log(CH3COO-/CH3COOH)

for CH3COOH, pKa = 4.75

mol of NaCH3COO = mass/MW = (8.203/82.04) = 0.09998

mol of CH3COOH = MV = (0.1)(1) = 0.1 mol

pH = 4.75 + log(0.09998/0.1) = 4.7499

b)

after adding 5 mL of 0.5 M of NaOH

mol of NaOH = MV = (5/1000)*0.5 = 0.0025 mol of OH-

after reactions:

mol of CH3COOH left = 0.1 - 0.0025 = 0.0975

mol of CH3COO- formed = 0.09998+0.0025 = 0.10248

so..

pH = pKa + log(CH3COO-/CH3COOH) = 4.75 + log(0.10248/0.0975) = 4.771

c)

can't be calculated if buffer B is not stated