Question

In lab, you will be provided with a 1.8 M acetic acid solution (CH3COOH) and solid sodium acetate trihydrate (NaOOCCH3  3H2O). You want to make 100.0 mL of a 0.5 M acetic acid buffer at pH 5.00. How many grams of sodium acetate trihydrate, and what volume (mL) of the 1.8M acetic acid solution and deionized water should you combine? The pKa of acetic acid is 4.75. Show all work. (6 pts)

Answer 1

Answer:

you will need 4.35 grams of Sodium acetate trihydrate and 10 ml of 1.8 M acetic acid solution.

Explanation:

Because we are dealing with a buffer solution, we can use Henderson-Haselbach equation: pH=pKa+log(A/HA)

HA = weak acid = acetic acid solution

A = conjugate base = Sodium acetate trihydrate

Step 1: find the ratio of A and HA

pH-pKa=log(A/HA) thus A/HA = 10^(pH-pKa)= 10^(5-4.75)=10^0.25=1.78

Step 2: Find % of each component

A/HA= 1.78/1 where total = 1.78+1=2.78

%A= 1.78/2.78 x 100 = 64%; %HA=1/2.78 x 100 = 36%

step 3: Amount of each in buffer solution

A= 64% of 0.5 M= 0.32 M; HA= 36% of 0.5 M = 0.18 M

step 4: grams of Sodium acetate trihydrate (Mm = 136.079 g/mole)

v=100 ml=0.1 L

mole=C×V=0.32×0.1=0.032 mole; mass=n×Mm=0.032×136.079=4.35 grams

Step 5: Volume of 1.8 M acetic acid solution

C1V1 = C2V2 where C1=1.8 M V1= what we are looking for; C2=0.18 M V2=100 ml

V1 = C2V2/C1

V1=(0.18×100)/1.8=10 ml