Buffers and the Henderson-Hasselbalch Equation
Problems #11 - 20

Problem #11: Part A: Which of the following buffer systems would be the best choice to create a buffer of pH = 7.20?

(a) CH₃COOH & CH₃COOK
(b) HClO₂ & KClO₂
(c) NH₃ & NH₄Cl
(d) HClO & KClO

Solution to A:

1) Examine the pK_a of each acid:

(a) K_a = 1.77 x 10¯⁵; pK_a = 4.752
(b) K_a = 1.1 x 10¯²; pK_a = 1.96
(c) K_a = 5.65 x 10¯¹⁰; pK_a = 9.248
(d) K_a = 2.9 x 10¯⁸; pK_a = 7.54

2) Write the Henderson-Hasselbalch equation:

pH = pK_a + log [base / acid]

3) Set the base/acid ratio to 1/1 and the result is:

pH = pK_a

4) Look for the pK_a value nearest to a pH of 7.20 when the base/acid ratio is 1/1:

the best choice for a pH = 7.20 buffer is answer choice d, the HClO & KClO buffer

5) Why?

The buffer with a pK_a closest to 7.20 would be the best choice because it has the least adjusting to do in the base/acid ratio in order to reach a buffer of pH = 7.20. I will return to this below.

Solution to B:

1) Let's write the Henderson-Hasselbalch equation again:

pH = pK_a + log [base / acid]

2) We know the desired pH of the buffer, so let's put it in:

7.20 = pK_a + log [base / acid]

3) Let's get the pK_a of HClO:

I will use a pK_a equal to 7.5376 (I'll keep a couple extra digits for the moment.)

4) Put it in place:

7.20 = 7.5376 + log [base / acid]

log [base / acid] = −0.3376

[base / acid] = 0.45962 (still keeping some extra digits)

5) The above ratio is a ratio of moles, but we want a ratio of weight. What we will do is use the molar masses of KClO and HClO and change the mole ratio to a weight ratio.

numerator (KClO) = (0.45962 mol) (90.55 g/mol) = 41.6186 g
denominator (HClO) = (1 mol) (52.4599 g/mol) = 52.46 g

The weight ratio is 41.6186 g / 52.46 g = 0.79 (to two sig figs)

Comment:

Using the KClO & HClO buffer is the best choice because it preserves the greatest buffering capacity. Let's see what weight ratio is needed to make a pH = 7.20 buffer with acetic acid and potassium acetate. (I'll use HAc and KAc as abbreviations.)

pH = pK_a + log [base / acid]

7.20 = 4.752 + log [base / acid]

log [base / acid] = 2.448

[base / acid] = 280.54

numerator (KAc) = (280.54 mol) (98.1417 g/mol) = 27532.67 g
denominator (HAc) = (1 mol) (60.0516 g/mol) = 60.05 g

The weight ratio is 27532.67 g / 60.05 g = 458.5

A buffer with 458.5 g of KAc for every one gram of HAc? Cute!

If you needed a buffer at pH = 4.00, then KAc/HAc is your best choice. It's not the best choice for pH = 7.20!

Problem #12: What is the [A¯]/[HA] ratio when a weak acid is in a solution with the pH one unit below its pK_a?

Solution:

1) Set pK_a to be K and the pH to be K − 1. Now, use the Henderson-Hasselbalch equation:

pH = pK_a + log ([A¯]/[HA])

2) Substitute:

K − 1 = K + log ([A¯]/[HA])

3) Move K to the other side:

−1 = log ([A¯]/[HA])

4) Antilog both sides:

0.1 = ([A¯]/[HA])

5) Replace 0.1 with a fraction:

1/10 = ([A¯]/[HA])

Problem #13: What is the pH of the resulting solution if 40.0 mL of 0.432 M methylamine, CH₃NH₂, is added to 15.0 mL of 0.234 M HCl? Assume that the volumes of the solutions are additive. pK_a = 10.66 for CH₃NH₃⁺ (Found here.)

Solution:

1) HCl is an acid and it will react with the base. In problems of this nature, the HCl is the limiting reagent and, at the end of the reaction, we have a solution with some weak base (the methylamine) and the salt of a weak base (the methylammonium chloride, CH₃NH₃Cl).

2) What we wind up with is a buffer. We will use the Henderson-Hasselbalch equation at the end to get the pH, but we need to get moles of the reactants first.

HCl ---> (0.234 mol/L) (0.0150 L) = 0.00351 mol
CH₃NH₂ ---> (0.432 mol/L) (0.0400 L) = 0.01728 mol

3) 100% of the HCl is used up. (If it wasn't, the solution would essentially be a solution of a strong acid. Calculation of that pH is not the goal in this problem.) What remains is this:

methylamine ---> 0.01728 mol − 0.00351 mol = 0.01377 mol
methylammonium ion ---> 0.00351 mol

4) Now, we use the H-H

pH = pK_a + log (base/acid)

pH = 10.66 + log [0.01377 / 0.00351] = 11.2536

11.12 to two sig figs. The driving force on the sig figs is the two sig figs of 10.66.

5) Look here:

log [0.01377/0.00351]

Notice how I used a ratio of moles and not of molarity. That s because both mole amounts are in the same volume of 55 mL. If you calculated the molarities and determined the ratio, the ratio of molarities would be the same as the ratio of moles.

Problem #14: Calculate the pH after 0.020 mol of HCl is added to 1.00 L of 0.10 M NaC₃H₅O₂. K_a = 1.3 x 10^-5.

Solution:

1) After adding the HCl, the solution will contain some propanoic acid and some sodium propionate (the NaC₃H₅O₂). The solution is a buffer and we will use the Henderson-Hasselbalch equation at the end to get the pH.

2) Be aware that, in these types of problems, the strong acid (or strong base) will ALWAYS be the limiting reagent. If it was in excess, the excess strong acid (or strong base) would dominate the pH and the solution would essentially be that of a strong acid (or a strong base).

3) But first, we require the moles of the two substance after the HCl reacts.

Before the HCl reacts:

NaC₃H₅O₂ ---> (0.10 mol/L) (1.0 L) = 0.10 mol
HCl ---> 0.02 mol <--- it's the limiting reagent

After the HCl reacts

NaC₃H₅O₂ ---> 0.10 mol minus 0.02 mol = 0.08 mol
HC₃H₅O₂ ---> 0.02 mol

4) We need the pK_a:

−log 1.3 x 10^-5 = 4.886

5) Now, the H-H

pH = pK_a + log (base/acid)

pH = 4.886 + log (0.08 / 0.02)

pH = 4.886 + 0.602 = 5.488 = 5.49

6) Notice that I just used moles in the ratio just above rather than molarity. The two substances involved were in 1.0 L, so the moles are numerically equal to the molarities. Even if it were in 500 mL (for example), using the moles rather than molarity would be just fine.

Problem #15: A buffer with a pH of 4.31 contains 0.31 M of sodium benzoate and 0.24 M of benzoic acid. What is the concentration of [H⁺] in the solution after the addition of 0.050 mol of HCl to a final volume of 1.3 L? Assume that any contribution of HCl to the volume is negligible.

Solution:

1) We can use the Henderson-Hasselbalch equation to solve this problem. Here's the H-H:

pH = pK_a + log [base/acid]

2) Notice that the problem does not provide you with the pK_a of benzoic acid. I think this is because the question writer (NOT the ChemTeam!) wants you to first calculate the pK_a from the data provided.

4.31 = pK_a + log (0.31 / 0.24)

pK_a = 4.20

3) We need to know how much benzoate is used up as we react it with the HCl.

(0.31 mol/L) (1.3 L) = 0.403 mol

0.403 mol − 0.050 mol = 0.353 mol

4) The amount of benzoic acid went up by the same 0.050 mol:

(0.24 mol/L) (1.3 L) = 0.312 mol

0.312 mol + 0.050 mol = 0.362 mol

5) We can divide the new moles by 1.3 to get molarities and then use the molarities in the H-H equation. Or, we can just use the mole amounts, since they are in the same ratio as the molarities.

pH = 4.20 + log (0.353 / 0.362)

pH = 4.20 + (−0.011)

pH = 4.189

6) Do this to get the [H⁺]:

[H⁺] = 10¯^pH = 10¯^4.189 = 0.000065 M

7) Note the approximate 0.12 change (4.31 to 4.19) in the pH of the buffer. Adding 0.050 mol of HCl to pure water takes the pH from 7 down to 1.4. Much bigger pH change!

Problem #16: What would the concentration of sodium formate (NaCOOH) be in 0.00750 M formate buffer at pH 4.358?

Solution:

1) We need to know the K_a of formic acid:

1.77 x 10¯⁴

2) The "0.00750 M formate buffer" references the total molarity of formic acid and sodium formate in the buffer. What are the concentrations of each?

formic acid: 0.00750 − x
sodium formate: x

3) Now, for the Henderson-Hasselbalch:

pH = pK_a + log [base / acid]

4.358 = 3.752 + log [x / (0.00750 − x)]

log [x / (0.00750 − x)] = 0.606

x / (0.00750 − x) = 4.03645

x = 0.030273 − 4.03645x

5.03645x = 0.0030273

x = 0.00601 M

4) We can check by putting values back into the H-H:

pH = 3.752 + log [0.00601 / 0.00149]

pH = 3.752 + 0.606

pH = 4.358

Problem #17: In 1.00 L of solution, 0.529 mole of HNO₂ is added to 0.246 mole of NaOH. (Nitrous acid has a K_a of 4.0 x 10¯⁴.) What is the final pH?

Solution:

1) Allow the HNO₂ to react with the NaOH to form some NaNO₂:

moles HNO₂ remaining ---> 0.529 − 0.246 = 0.283 mol
moles NaNO₂ formed ---> 0.246 mol

HNO₂ and NaOH react in a 1:1 molar ratio and form NaNO₂ in a 1:1 ratio.

2) Use the Henderson-Hasselbalch Equation:

pH = 3.40 + log (0.246 / 0.283]

pH = 3.40 + (−0.06)

pH = 3.34

Problem #18: You desire to create a solution with a pH of 3.26. If you add 0.577 moles of HF to 1.00 L of solution, how many moles of NaF should you add?

Solution:

1) We look up the K_a of HF:

7.2 x 10¯⁴

2) Use the Henderson-Hasselbalch Equation:

3.26 = 3.14 + log (x / 0.577)

log (x / 0.577) = 0.12

x / 0.577 = 1.318

x = 0.76 mol

Problem #19: If you begin with 48.2 mL of a 0.171 M solution of HNO₂, how many grams of NaNO₂ would you have to add to the solution for a pH of 3.32?

Solution:

Note: we will assume that the addition of solid NaNO₂ to the solution will not cause a volume change.

1) Use the Henderson-Hasselbalch Equation:

pH = pK_a + log (base / acid)

3.32 = 3.40 + log (x / 0.171)

log (x / 0.171) = −0.08

x / 0.171 = 0.83176

x = 0.14223 M <--- I'll keep some guard digits

2) Determine grams of NaNO₂ required to make 48.2 mL of a 0.14223 M solution:

MV = mass / molar mass

(0.14223 mol/L) (0.0482 L) = mass / 68.995 g/mol

mass = 0.473 g

Problem #20: Calculate the ratio of the concentration of acetic acid and acetate required in a buffer system at a pH of 4.208 (the pK_a of acetic acid equals 4.752).

Solution:

pH = pK_a + log [base / acid]

4.208 = 4.752 + log [base / acid]

log [base / acid] = −0.544

base / acid = 0.285759

Therefore:

acid / base ratio is 3.50 to 1

Bonus Problem: You need to prepare a buffer solution of pH 3.972 from 10.0 mL of 0.335 M solution of a weak acid whose pK_a is 3.843. What volume of 0.385 M NaOH would you need to add?

Solution:

1) Use the H-H equation to calculate the ratio of conjugate base to acid in the desired buffer:

	[A¯]
pH = pKa + log	–––––
	[HA]

	[A¯]
3.972 = 3.843 + log	–––––
	[HA]

	[A¯]
0.129 = log	–––––
	[HA]

	[A¯]
1.346 =	–––––
	[HA]

[A¯] = 1.346[HA]

2) We also know that [A¯] + [HA] = 0.335 M. This is an important point that calls for a bit of explanation:

Regardless of how much NaOH is added, the changes in the amounts of HA and A¯ are directly related to each other. In other words:

[A¯] + [HA] = 0.335 M

Why is this? Conservation of mass requires that atoms of A in the solution can not be created and can not be destroyed. These atoms simply move being part of aqueous HA or part of the aqueous ion A¯. This means the sum of the moles of HA and A¯ must remain constant. Since these two species always share the same volume, the sum of their molarities also must remain constant.

3) Substituting the previous equation gives:

1.346[HA] + [HA] = 0.335

[HA] = 0.143 M and by subtraction, [A¯] = 0.192 M

4) Calculate moles of A¯. This will equal moles of NaOH needed:

(0.192 mol/L) (0.010 L) = 1.92 x 10¯³ mol A¯ = 1.92 x 10¯³ mol OH¯

5) Calculate the volume of NaOH required:

1.92 x 10¯³ mol / 0.385 mol/L = 4.99 x 10¯³ L = 4.99 mL NaOH