### Buffers and the Henderson-Hasselbalch EquationProblems #1 - 10

Problem #1: 20.0 cm3 of 1.00 mol HNO2 and 40.0 cm3 of 0.500 NaNO2 are mixed. What is the pH of the resulting solution? pKa of nitrous acid is 3.34

Comment: this is an answer that does not mention the Henderson-Hasselbalch Equation. In one's travels, one occasionally runs across an individual that does not like the H-H and insists on using the Ka expression. Just be aware that the Ka expression and the Henderson-Hasselbalch Equation are the same thing.

Solution:

1) Common ion effect....

The presence of a common ion, in this case the nitrite ion, will reduce the ionization of the weak acid HNO2 (which is only observed in aqueous solution.)

2) The HNO2 is diluted by the addition of NaNO2 solution.

20 mL HNO2 x 1.0vM / 60 mL = 0.33vM HNO2

3) The NO2¯ concentration is 2/3 of its original concentration.

2/3 x 0.5 = 0.33M NO2¯

4) Since the HNO2 and NO2¯ concentrations are the same, they will cancel out leaving the hydrogen ion concentration equal to the Ka, and the pH of the solution equal to the pKa.

Ka = [H+][NO2¯] / [HNO2]

4.57 x 10¯4 = [(x)(0.33M)] / 0.33M

x = 4.57 x 10¯4 = [H+]

5) pH = -log[H+]

pH = -log(4.57x10¯4)

pH = 3.34

Problem #2: Calculate the pH of a solution prepared by mixing 15.0 mL of 0.100 M NaOH and 30.0 mL of 0.100 M benzoic acid soluion. (Benzoic acid is monoprotic; its dissociation constant is 6.46 x 10¯5.)

Solution:

Moles of NaOH = (0.100 mol/L) (0.0150 L) = 0.001500 mol
Moles of benzoic acid = (0.100 mol/L) (0.0300 L) = 0.00300 mol

NaOH and benzoic acid react in a 1:1 molar ratio. After reaction, we have:

moles of sodium benzoate = 0.00150 mol
moles of benzoic acid = 0.00150 mol

pH = pKa + log [base / acid]

pH = 4.190 + log [0.00150 / 0.00150]

pH = 4.190

Problem #3: Calculate the pH of a solution prepared by mixing 5.00 mL of 0.500 M NaOH and 20.0 mL of 0.500 M benzoic acid solution. (Benzoic acid is monoprotic: its ionization constant is 6.46 x 10¯5.)

Moles of NaOH = (0.500 mol/L) (0.00500 L) = 0.00250 mol
Moles of benzoic acid = (0.500 mol/L) (0.0200 L) = 0.00750 mol

NaOH and benzoic acid react in a 1:1 molar ratio. After reaction, we have:

moles of sodium benzoate = 0.00250 mol
moles of benzoic acid = 0.00750 mol

pH = pKa + log [base / acid]

pH = 4.190 + log [0.00250 / 0.00750]

pH = 4.190 + (-0.477)

pH = 3.713

Problem #4: How many grams of NH4Cl need to be added to 1.50 L of 0.400 M ammonia in order to make a buffer solution with pH of 8.58? Kb for ammonia is 1.77 x 10¯5

Solution:

1) Use the Henderson-Hasselbalch Equation to solve this problem:

pH = pKa + log [base / acid]

2) I already know the pH, so:

8.58 = pKa + log [base / acid]

3) I need a pKa, not a Kb or a pKb. What I need is the pKa of the ammonium ion. I get it from Kw and the Kb of ammonia:

Kw = KaKb

1.00 x 10¯14 = (Ka) (1.77 x 10¯5)

Ka = 5.6497 x 10¯10

pKa = - log 5.6497 x 10¯10 = 9.24797 (I'll carry some extra digits.)

8.58 = 9.24797 + log [base / acid]

4) Next are the moles of NH3:

(0.400 mol/L) (1.50 L) = 0.600 mol

8.58 = 9.24797 + log [0.6 / x]

The acid (which is the NH4Cl, specifically the ammonium part) is our unknown. Note: at the end of my solution is an alternate solution path that solves for the required molarity of the NH4Cl. For there you can get to the required grams very easily.

5) Solve:

8.58 = 9.24797 + log [0.6 / x]

log [0.6 / x] = -0.66797

0.6 / x = 0.2147979

x = 2.7933234 mol

2.7933234 mol times 53.4916 g/mol = 149.42 g

to three sig figs, 149 g

6) We can check this by going back to the H-H Eq (and I will use molarities):

pH = 9.24797 + log [base / acid]

pH = 9.24797 + log [0.400 / 1.8622156]

pH = 9.24797 + log 0.2147979

pH = 9.24797 + (-0.66797)

pH = 8.58

The molarity of the NH4Cl came from this calc:

2.7933234 mol / 1.50 L

Problem #5: How many grams of dry NH4Cl need to be added to 2.40 L of a 0.800 M solution of ammonia to prepare a buffer solution that has a pH of 8.90? Kb for ammonia is 1.77 x 10¯5.

Solution:

pH = pKa + log [base / acid]

8.90 = 9.248 + log [0.800 / x]

log [0.800 / x] = -0.348

[0.800 / x] = 0.4487454

x = 1.782748 M (this is the required molarity of the NH4Cl

(1.782748 mol/L) (2.40 L) = mass / 53.4916 g/mol

mass = 228.8689 g

To three sig figs, 229 g

Problem #6: Determine the pH of the buffer made by mixing 0.0300 mol HCl with 0.0500 mol CH3COONa in 2.00 L of solution. The Ka of acetic acid is 1.77 x 10¯5.

Solution:

1) The hydrogen ion from the HCl will protonate the acetate ion according to this reaction:

H+ + CH3COO¯ ---> CH3COOH

The key point is that the reaction has a 1:1 molar ratio between the two reactants.

2) The HCl is the limiting reagent. Some CH3COOH will be made and some CH3COO¯ will be left over.

CH3COOH ---> 0.0300 mol is made (as a result of H+ + CH3COO¯)
CH3COO¯ ---> 0.0500 - 0.0300 = 0.0200 mol remains

3) The Henderson-Hasselbalch Equation is used to determine the pH:

pH = pKa + log [base / acid]

pH = 4.752 + log [0.0200 / 0.0300]

pH = 4.752 + 0.176 = 4.928

Problem #7: Determine the pH of the solution after the addition of 133 mL of 0.0270 M nitric acid (HNO3) to 172 mL of 0.0570 M sodium hydrogen citrate (Na2C3H5O(COOH)(COO)2)

Solution:

N.B. The presence of the sodium ion will be ignored since it is a spectator ion.

1) The nitric acid is a strong acid and it will protonate the weaker acid, symbolized by HCit2¯. This is the reaction:

HCit2¯ + H+- ---> H2Cit¯ <--- this reaction goes to completion, not an equilibrium

2) If you flip the above reaction, you have the equation for the Ka2 of citric acid:

H2Cit¯ ⇌ HCit2¯ + H+ <--- note that this reaction is an equilibrium

3) We need to know how much HCit2¯ reacted and how much H2Cit¯ was made:

moles HNO3 ---> (0.0270 mol/L) (0.133 L) = 0.003591 mol
moles HCit2¯ ---> (0.0570 mol/L) (0.172 L) = 0.009804

How much H2Cit¯ was made? Ans: 0.003591 mol <--- all the H+- from the nitric acid was used up
How much HCit2¯ remains? Ans: 0.009804 - 0.003591 = 0.006213 mol

The Ka2 for citric acid is 1.73 x 10¯5. I found it here.

4) We now use the Henderson-Hasselbalch equation since we have a buffer:

pH = pKa + log [base / acid]

pH = pKa + log (HCit2¯ / H2Cit¯)

Note: the HCit2¯ is the base since it lacks the proton. The H2Cit¯ is the acid since it has the proton that gets donated to a water molecule to make H3O+ (signified in the equation in step 2 by H+).

pH = 4.762 + log (0.006213 / 0.003591)

pH = 4.762 + 0.238

pH = 5.000

Problem #8: A buffer with a pH of 4.30 contains 0.33 M of sodium benzoate and 0.26 M of benzoic acid. What is the concentration of H+ in the solution after the addition of 0.058 mol of HCl to a final volume of 1.6 L?

Solution:

1) Determine pKa:

4.30 = pKa + log(0.33/0.26)

4.30 = pKa + 0.104

pKa = 4.196

You can also look up pKa (or Ka) online. I decided to calculate it.

2) Determine moles after addition of HCl:

0.330 - 0.058 = 0.272 mole sodium benzoate

0.260 + 0.058 = 0.318 mole benzoic acid

3) Determine new pH:

pH = 4.196 + log(0.272/0.318)

pH = 4.196 - 0.068

pH = 4.128

4) Convert pH to [H+]:

[H+] = 10-4.128 = 7.745 x 10-5 M

Problem #9: A buffer solution contains 5.00 mL of 2.00 M acetic acid, 45.0 mL water and 2.05 g sodium acetate. Predict the pH of the buffer solution.

Solution:

The final volume of buffer is 50 mL (5 mL acid + 45 mL water). We'll ignore significant figures until the end.

1) Calculate the molarity of the final acid solution:

M1V1 = M2V2

(M1) (50 mL) = (2.0 mol/L) (5 mL)

M1 = 0.2 M

2) Calculate the molarity of the CH3COONa (molar mass = 82.03 g/mol):

moles in 2.05 g = 2.05/82.03 = 0.0249 mole in 0.050 L solution

molarity = 0.0249 mol / 0.05 L = 0.498 M

3) Now apply the Henderson-Hasselbalch equation:

pH = pKa + log ([base] / [acid])

The pKa of CH3COOH (not sodium acetate) is 4.752

pH = 4.752 + log (0.498/0.20)

pH = 4.752 + log 2.49

pH = 4.752 + 0.396

pH = 5.148

4) You can also solve this problem by determining the number of moles of acetic acid and the number of moles of sodium acetate:

acetic acid ---> (2.0 mol/L) (0.005 L) = 0.01 mol

pH = 4.752 + log (0.0249 / 0.01)

Problem #10: What volumes of 0.100 M acetic acid and 0.100 M sodium acetate would be required to produce 1.00 L of buffer at pH 4.000? (pKa = 4.752)

Solution:

pH = pKa + log [Ac¯]/[HAc]

4.000 = 4.752 + log (Ac¯/HAc)

log Ac¯/HAc = -0.752

log HAc/Ac¯ = 0.752

N.B. an acidic pH means we'll need more acid than base, so flip the ratio to put the larger value on top.

[HAc]/[Ac¯] = 5.64937 <--- carry a couple guard digits

This means you need 5.64937 times as much HAc as Ac¯

To make 1000 mL:

5.64937x + x = 1000

6.64937x = 1000

x = 150.39 mL of acetate

plus 849.61 mL acetic acid

Bonus Problem: A solution containing 0.0158 M a diprotic acid with the formula H2A and 0.0226 M of its salt Na2A. The K2 values for the acid are 1.20 x 10¯2 (Ka2) and 5.37 × 10¯7 (Ka2). What is the pH of the solution?

Solution:

1) Write the two Ka equations:

 H2A ⇌ HA¯ + H+ Ka1 HA¯ ⇌ A2¯ + H+ Ka2

2) Reverse the second equilibrium:

 H2A ⇌ HA¯ + H+ Ka1 A2¯ + H+ ⇌ HA¯ 1/Ka2

3) Add the two chemical equations:

 H2A + A2¯ ⇌ 2HA¯ K' = Ka1 / Ka2

4) Solve for K':

K' = 1.20 x 10¯2 / 5.37 x 10¯7 = 22346

What that means is that there is a VERY product (the HA¯) favored equilibrium.

5) What we now do is treat this as a limiting reagent problem:

H2A + A2¯ ⇌ 2HA¯

Assume 1.00 L of solution is present.

moles H2A ---> 0.0158
moles A2¯ ---> 0.0226

The 1:1 stoichiometry of H2A reacting with A2¯ means H2A (the lower amount at 0.0158) is the limiting reagent.

6) Determine amount of HA¯ and H2A in solution:

A2¯ ---> 0.0226 - 0.0158 = 0.0068
HA¯ ---> (2) (0.0158) = 0.0316

Reaction stoichiometry: two HA¯ produced for every one H2A used.

7) Calculate the pH as a buffer centered around Ka2

pH = 6.270 + log (0.0068/0.0316) = 5.603