Sample pH / Buffer Problems

 

Common Biochemical Buffers

Buffer System

MW

pKa1

pKa2

pKa3

Formic acid (HCOOH)

46.02

3.75

 

 

Acetic Acid (CH3COOH)

60.04

4.76

 

 

Phosphoric (H3PO4)

97.97

2.2

6.86

12.4

Carbonic (H2CO3)

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) Tris-HCl 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 TrisH, need to add 0.326 liters of 1 M HCl

 


(d)  
Add H2O 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
(ml)

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+] = (.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