BUFFER SOLUTION ASSIGNMENT

pH Changes in Buffered and Unbuffered Solutions
Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might change the biochemical activity of these compounds.

(a) Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.

Solution
To determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples):

1. Determine the direction of change. The equilibrium in a mixture of H₃O⁺, , and CH₃CO₂H is:
   

   The equilibrium constant for CH₃CO₂H is not given, so we look it up in : K_a = 1.8 × 10⁻⁵. With [CH₃CO₂H] =  = 0.10 M and [H₃O⁺] = ~0 M, the reaction shifts to the right to form H₃O⁺.

2. Determine x and equilibrium concentrations. A table of changes and concentrations follows: 
3. Solve for x and the equilibrium concentrations. We find:
   

   and

   

   Thus:

   

   
4. Check the work. If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, Q = K_a.

(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer, giving a solution with a volume of 101 mL.

First, we calculate the concentrations of an intermediate mixture resulting from the complete reaction between the acid in the buffer and the added base. Then we determine the concentrations of the mixture at the new equilibrium:

1. Determine the moles of NaOH. One milliliter (0.0010 L) of 0.10 M NaOH contains:
   
2. Determine the moles of CH₂CO₂H. Before reaction, 0.100 L of the buffer solution contains:
   
3. Solve for the amount of NaCH₃CO₂ produced. The 1.0 × 10⁻⁴ mol of NaOH neutralizes 1.0 × 10⁻⁴ mol of CH₃CO₂H, leaving:
   

   and producing 1.0 × 10⁻⁴ mol of NaCH₃CO₂. This makes a total of:

   
4. Find the molarity of the products. After reaction, CH₃CO₂H and NaCH₃CO₂ are contained in 101 mL of the intermediate solution, so:
   
   

   Now we calculate the pH after the intermediate solution, which is 0.098 M in CH₃CO₂H and 0.100 M in NaCH₃CO₂, comes to equilibrium. The calculation is very similar to that in part (a) of this example:

   

   This series of calculations gives a pH = 4.75. Thus the addition of the base barely changes the pH of the solution

(c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74 (a 1.8 × 10⁻⁵–M solution of HCl). The volume of the final solution is 101 mL.

Solution
This 1.8 × 10⁻⁵–M solution of HCl has the same hydronium ion concentration as the 0.10-M solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains:

As shown in part (b), 1 mL of 0.10 M NaOH contains 1.0 × 10⁻⁴ mol of NaOH. When the NaOH and HCl solutions are mixed, the HCl is the limiting reagent in the reaction. All of the HCl reacts, and the amount of NaOH that remains is:

The concentration of NaOH is:

The pOH of this solution is:

The pH is:

The pH changes from 4.74 to 10.99 in this unbuffered solution. This compares to the change of 4.74 to 4.75 that occurred when the same amount of NaOH was added to the buffered solution described in part (b).

Check Your Learning
Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8 × 10⁻⁵MHCl solution from 4.74 to 3.00.

Answer:

Initial pH of 1.8 × 10⁻⁵M HCl; pH = −log[H₃O⁺] = −log[1.8 × 10⁻⁵] = 4.74

Moles of H₃O⁺ in 100 mL 1.8 × 10⁻⁵M HCl; 1.8 × 10⁻⁵ moles/L × 0.100 L = 1.8 × 10⁻⁶

Moles of H₃O⁺ added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L × 0.0010 L = 1.0 × 10⁻⁴ moles; final pH after addition of 1.0 mL of 0.10 M HCl:

This set of Biochemistry Questions and Answers for Freshers focuses on “Buffering against pH changes in Biological Systems”.

1. Which can act as buffer?
a) NH₄ Cl + HCl
b) CH₃ COOH + H₂ CO₃
c) 40ml of 0.1M NaCN + 20ml of 0.1M HCN
d) NaCl + NaOH
View Answer

2. Calculate the pH of a mixture of 0.10M acetic acid and 0.20M sodium acetate. The pK_a of acetic acid is 4.76.
a) 5.1
b) 4.1
c) 6.1
d) 7.1
View Answer

3. Calculate the pK_a of lactic acid, given that when the concentration of lactic acid is 0.010M and the concentration of lactate is 0.087M, the pH is 4.80.
a) 4.0
b) 3.9
c) 3.3
d) 4.1
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4. Calculate the ratio of the concentrations of acetate and acetic acid required in a buffer system of pH 5.30.
a) 3.2
b) 3.3
c) 3.4
d) 3.5
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5. Assertion A: pH of a buffer solution solution does not change on dilution.
Reason R: On dilution the ration of concentration of salt and acid (or base) remains unchanged.
a) A and R are true, R is the correct explanation of A
b) A and R are true, R is not correct explanation of A
c) A is true but R is false
d) A is false but R is true
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6. Which of the following mixture in aqueous solution of equimolar concentration acts as a buffer solution?
a) HNO₃ + NaOH
b) H₂ SO₄ + KOH
c) NH₄ OH(excess) + HCl
d) CH₃ COOH + NaOH(excess)
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7. 1M NaCl and 1M HCl are present in an aqueous solution. The solution is
a) Not a buffer solution with pH < 7
b) Not a buffer solution with pH > 7
c) A buffer solution with pH < 7
d) A buffer solution with pH > 7
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8. For an acid buffer solution the pH is 3. The pH can be increased by
a) Increasing the concentration of salt
b) Increasing the concentration of acid
c) Decreasing the concentration of salt
d) Independent of concentration of acid & salt
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9. The buffer capacity is equal to
a) Δn / ΔpH
b) pH / Δn
c) ± 1pKa
d) ± 2pKa
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10. Buffer capacity is maximum when
a) One mole of NH₄ Cl is added to two moles of NH₄ OH
b) One mole of NH₄ Cl is added to one moles of NH₄ OH
c) One mole of NH₄ Cl is added to one mole of NaOH
d) One mole of NaCl is added to one mole of NaOH
View Answer