Adrian Ramirez
Pranav Veera
Buffers
Part B:
1-) Formic Acid and Sodium Formate (Target pH=3.74)
log[A/B]=pH-pKa
log[A/B] =3.74-3.74
[A/B]=0
A/B = 1
0.1M=[A]+[B]
[A]=[B]=0.05M
M1V1= M2V2
(0.50M) V1= (0.05)x(0.1)
V1=0.01L= 10ml (of Formic Acid and Sodium Formate needed)
2-) Formic Acid and Sodium Formate (Target pH=4.04)
log[A/B]=pH-pKa
log[A/B]=3.74-4.04
[A]/[B]=10-0.3=0.501
[B]=(0.501) x[A]
0.1M=[A]+[B]
0.1M= [A]+(0.501) A
0.1M=1.501A
A=0.067; B=0.033M
M1V1=M2V2
(0.5M) V1= (0.067) (100)
V1=13.4mL
(0.5) V1= (0.033) (100ml)
V1= 6.6mL
Acetic acid and sodium acetate (target pH= 4.44 and 5.04) for buffer 3 and 4.
Part C: Determination of buffer capacity
Acid/Base Pair
Formic acid/
Formic acid/
Acetic acid/
Acetic acid/
Sodium formate Sodium formate 0.005 0.0033
Sodium acetate 0.0067
Sodium acetate 0.0033
Mols of Acid in
Buffer Target pH Initial pH Vol of NaOH Mol of NaOH
3.74 3.84 9.53 0.0368
4.04 3.7 4.7 0.00235
4.44 4.45 5.0 0.0025
5.04 5.05 1.8 0.0009
The sodium formate and formic acid; buffer 1 and 2, the initial pH was not very close to the
target pH. Buffer 1 initial pH has a margin of error of 10%; while buffer 2 has a margin of error
of almost 34%. In contrast, in buffer of acetic acid and sodium acetate the initial pH was almost
the target pH; with a margin of error of about 1%. Some possible causes of this elevated
percentage of error may be due to measurement and observational errors; also, it could be due to
contamination or environmental influences. In order to reach buffer capacity, we added between
0.0009 and 0.0368 mol of NaOH, this is because NaOH is a strong base thus it will react with the
acid present in the buffer. I can conclude that Acetic acid/ Sodium acetate has a greater buffer
capacity because it took most of the excess base taking the initial pH closer to the target pH.
References:
1-https://www.voutube.com/watch?v=HzFIdpThd-s
2-The Properties of Buffer Handou