Sample pH /
Buffer Problems
Common
Biochemical Buffers
Buffer System 
MW 
pKa1 
pKa2 
pKa3 
Formic acid (HCOOH) 
46.02 
3.75 


Acetic Acid (CH_{3}COOH) 
60.04 
4.76 


Phosphoric (H_{3}PO_{4}) 
97.97 
2.2 
6.86 
12.4 
Carbonic (H_{2}CO_{3}) 
62.01 
6.10 
10.22 

Tris (acid form) 
121.1 
8.15 


(These are for your practice. If you can do these, you should be able to do any I give you on the exams.)
1) Starting with solid Tris base and 1M HCl, describe how one would make 2 liters of 200 mM (= 0.2 M) TrisHCl buffer pH 7.5
(a) We want 2 liters of 0.2 M Tris, MW=121.1
(b) Calc acid / base ratio
(c)
To get .326 mol of Tris·H, need to add 0.326 liters of 1 M HCl
(d)
Add H_{2}O to 2000 ml
2) Starting with 5 M Phosphoric acid and 10 M NaOH, make 1000 ml of .25 M Na Phosphate buffer pH 7.
(a) Have 5 M H3PO4; want 0.25 M
(1 liter x 0.25
mol/liter)/(5 mol/liter) = 0.05 liter
(b) For pH 7, am concerned with pKa 2, so have to add enough
base to convert all the H3PO4 to H2PO4
(1 liter x 0.25
mol/liter)/(10 mol/liter) = 0.025 liter
(c) Add enough more base to get correct pH
pH = pKa + log
[base]/[acid]
7 = 6.86 + log
[base]/[acid]
1.38 =
[base]/[acid] ; therefore [base] = 1.38 [acid] and 2.38 [acid] = .25 mol/liter
[acid] = .105 M;
[base] = .145 M
Need to add additional .145 M of base
(1 liter x 0.145
mol/liter)/(10 mol/liter) = 0.0145 liter
SO: Mix 0.05 liters
5 M H3PO4, .0.0395 liters 10 M NaOH; add water to 1 liter.
3) Starting with 5 M Phosphoric acid and 10 M NaOH, make 1000 ml of .25 M Na Phosphate buffer pH 12.
Same as above, except use pKa3.
Add 0.05 liters 5 M H3PO4, .0571 liters 10M NaOH; add water
to 1 liter.
4) I have 500 ml of 0.2 M Na acetate buffer pH 4.0. To this I add 500 ml of 0.2 M Na acetate (not a buffered solution, just plain old sodium acetate). What is the pH of the resulting solution? (Nowhere near as hard as it looks at first glance.)
Solution 1 (0.2M NaAc pH4) contains the following:
[Ac] + [HAc] =
0.2 M
4 = 4.75 + log
[Ac]/[HAc]; 0.178 = [Ac]/[HAc];
1.178[HAc] = 0.2 M
[HAc] = .170 M;
0.170 mol/liter x .50 liter = 0.085 mol HAc
[Ac] = .030 M;
0.030 mol/liter x .50 liter = 0.015 mol Ac
Solution 2 contains:
.2 mol/liter x
0.5 liter = 0.1 mol Ac
After mixing:
Volume = 1 liter,
containing
0.085 mol HAc
0.100 + 0.014 mol
Ac
pH = 4.76 + log (.115 M)/(.085 M)
pH = 4.89
5) Draw the titration curve for the titration of 200 ml .05 M formic acid with 1 M potassium hydroxide.
KOH added 
pH 
0 
2.54 
1 
2.80 
2 
3.15 
3 
3.38 
4 
3.57 
5 
3.75 
6 
3.93 
7 
4.12 
8 
4.35 
9 
4.70 
9.5 
5.03 
9.9 
5.75 
6) What is the pH of each of the following solutions:
(a) .0.4 M phosphoric acid
Applicable pKa is pKa1 = 2.2
Ka = 10**(pKa) = 10**2.2 = .00631
Ka = ([H+]^2)/[H2PO4]
[H+] = Ö(.04)(.00631) = .0159
pH = log[H+] = 1.8
(b) mixture of 100 ml 1 M acetic acid and 10 ml 1 M Na acetate
pH = pKa + log(.01/.1) = 4.76 +
(1) = 3.76
(c) mixture 0f 10 ml 1 M acetic acid and 100 ml 1 M Na acetate
pH = pKa + log(.1/.01) = 4.76 + (1)
= 5.76
(d) .02 M HCl
HCl is a strong acid
PH = log[H+] = log(.02) = 1.70
(e) mixture of 100 ml 1 M potassium formate acid and 30 ml 1 M formic acid
pH = 3.75 + log (.1/.03) = 4.27