### Buffers and the Henderson-Hasselbalch Equation:Fifteen Examples

As a reminder, here is the Henderson-Hasselbalch Equation:

 [base] pH = pKa + log ––––– [acid]

Be aware that this:

pH = pKa + log [base / acid]

is often the way you see it written on the Internet. Many Q&A forums still lack the ability to make a more typeset-appearing H-H Equation.

Note: the bonus problem at the end of the file involves having to calculate how much of one of the buffer components is consumed and how much of the other is produced. There will be an unknown in the log portion of the Henderson-Hasselbalch. The third set of problems (#21 to 30) has more examples of this type.

A colleague has written a Deep Dive into the first three examples solved just below. In his document, he goes into detail concerning the approximations that are used to develop the Henderson-Hasselbalch Equation and why those approximations are valid (hint: the 5% rule). At the end, he details a case where the approximation fails and an exact calculation must be made.

The ChemTeam recommends you become familiar with H-H calculations, then return to the Deep Dive and give it a careful read. Knowledge of the inner workings of the H-H Equation will be to your chemical benefit.

Example #1: A buffer is prepared containing 1.00 M acetic acid and 1.00 M sodium acetate. What is its pH?

Solution:

1) To solve the above example, we must know the pKa of acetic acid. Often, the problem will provide the pKa. If the problem provides the Ka, you must convert it to the pKa (see below).

Comment: be aware, your teacher may create a test question where you must look up the Ka. Even in this era of fairly easy Internet access, try one of the appendices of your textbook. Tables of Ka values are also widespread on the Internet..

The Ka of acetic acid is 1.77 x 10¯5

2) Calculate the pKa:

pKa = −log Ka = −log 1.77 x 10¯5 = 4.752

3) Next, we simply insert the appropriate values into the H-H equation:

 1.00 pH = 4.752 + log ––––– 1.00

Since the log of 1 is zero, we have pH = 4.752

Comment: 1.8 x 10¯5 is a commonly-seen value for the Ka of acetic acid.

Notice that the pH is greater than the solution of just the pure acid (4.752 as compared to 2.376). This is due to LeChatelier's Principle. Consider the dissociation equation for acetic acid:
HAc ⇌ H+ + Ac¯

Increasing the concentration of the acetate (Ac¯) will push the equiibrium back to the left, decreasing the concentration of H+. This makes the solution less acidic, making the pH of the buffer larger than the pure acid solution.

Example #2: A buffer is prepared containing 0.800 molar acetic acid and 1.00 molar sodium acetate. What is its pH?

Solution:

The Henderson-Hasselbalch Equation:

 1.00 pH = 4.752 + log ––––– 0.800

x = 4.752 + 0.097 = 4.849

Note how decreasing the amount of acid makes the buffer pH become more basic (compare to example #1).

Example #3: A buffer is prepared containing 1.00 molar acetic acid and 0.800 molar sodium acetate. What is its pH?

Solution:

The Henderson-Hasselbalch Equation (done in the Internet way):

pH = pKa + log [base / acid]

x = 4.752 + log (0.800 / 1.00)

x = 4.752 − 0.097 = 4.655

Note how decreasing the amount of base makes the buffer pH become more acidic (compare to example #1).

Example #4: (a) Calculate the pH of a 0.500 L buffer solution composed of 0.700 M formic acid (HCOOH, Ka = 1.77 x 10¯4) and 0.500 M sodium formate (HCOONa). (b) Calculate the pH after adding 50.0 mL of a 1.00 M NaOH solution.

Solution to (a):

We can use the given molarities in the Henderson-Hasselbalch Equation:

pH = pKa + log [base / acid]

pH = 3.752 + log [0.500 / 0.700]

pH = 3.752 + (−0.146)

pH = 3.606

Solution to (b):

1) We need to determine the moles of formic acid and sodium formate after the NaOH was added. We first calculate the amounts before the addition of the NaOH:

HCOOH ---> (0.700 mol/L) (0.500 L) = 0.350 mol
HCOONa ---> (0.500 mol/L) (0.500 L) = 0.250 mol

2) Now, determine the moles of NaOH:

NaOH ---> (1.00 mol/L) (0.0500 L) = 0.0500 mol

3) NaOH reacts in a 1:1 molar ratio with HCOOH:

HCOOH ---> 0.350 mol − 0.0500 mol = 0.300 mol
HCOONa ---> 0.250 mol + 0.0500 mol = 0.300 mol

4) Calculate the new pH:

pH = 3.752 + log [0.300 / 0.300] <--- notice moles (not molarities) being used

pH = 3.752 + log 1

pH = 3.752

Part (b) of the above question is a popular one to ask. Be sure to know how to calculate the pH of a buffer after some strong acid or base has been added.

Note that you could add some weak acid or some weak base to a buffer. For example, adding ammonia (a weak base) to a buffer consisting on acetic acid and sodium acetate. The calculations for that type of situation are more complex and will not be addressed by the ChemTeam.

In other words, the only types of buffer calculation you will see in the ChemTeam is:

(a) a strong acid (or base) is added to an already-buffered solution
(b) a strong acid (or base) is added to a solution with just a weak base (or weak acid) in it, creating a buffer.

The reason moles cn be used is that the moles are in the same ratio as the molarities. To get the molarities, you would divide both moles by the same value. Remember, both the acid and the base in a buffer are in the same solution, so just one volume (used twice) is involved in changing moles to molarities.

Example #5: 0.1 mole of CH3NH2 (Kb = 5 x 10¯4) is mixed with 0.08 mole of HCl and diluted to one liter. What will be the H+ concentration?

Solution:

1) The reaction is this:

CH3NH2 + H+ ---> CH3NH3+

2) When 0.1 mole of CH3NH2 and 0.08 mole of HCl react, this is what remains after the reaction:

0.02 mol CH3NH2
0.08 mol CH3NH3+

3) Since we now have a buffer, we will use the Henderson-Hasselbalch Equation:

First, however, we will use the Kb to get the pKa.

pKa + pKb = 14

pKb = −log 5 x 10¯4 = 3.30103

pKa = 14 − 3.30103 = 10.69897

4) Now, use the H-H Equation:

pH = 10.69897 + log (0.02 / 0.08) <--- since it's in 1 liter, these are the molarities

pH = 10.69897 + (−0.60206)

pH = 10.09691

5) Now, we antilog the pH to get the H+ concentration:

[H+] = 10¯pH = 10¯10.09691 = 8 x 10¯11 M

Example #6: Calculate the pH when 25.0 mL of 0.200 M acetic acid is mixed with 35.0 mL of 0.100 M NaOH.

Solution:

1) Determine moles of each substance:

(0.200 mol/L) (0.0250 L) = 0.00500 mol of acetic acid
(0.100 mol/L) (0.0350 L) = 0.00350 mol of NaOH

2) Acetic acid and NaOH react in a 1:1 molar ratio. Determine the moles remaining after reaction (acetic acid is in excess):

acetic acid ---> 0.00500 mol − 0.00350 mol = 0.00150 mol

The acetic acid that reacts with the NaOH produces sodium acetate. In the solution will be 0.00350 mol of acetate anion (we may ignore the sodium ion. It plays no role in the pH.)

3) Use the Henderson-Hasselbalch equation to determine the pH:

pH = 4.752 + log (0.00350 / 0.00150)

pH = 4.752 + 0.368

pH = 5.120

Note that we did not have a buffer to begin with. There was a solution of acetic acid and some strong base was added resulting in a solution of a weak acid and its salt. In other words, a buffer.

Example #7: Calculate the pH when 50.0 mL of 0.180 M NH3 is mixed with 5.00 mL of 0.360 M HBr. (The Kb of ammonia is 1.77 x 10¯5.)

Solution:

1) Determine moles of each substance:

(0.180 mol/L) (0.0500 L) = 0.00900 mol of ammonia
(0.360 mol/L) (0.0050 L) = 0.00180 mol of HBr

2) Ammonia and HBr react in a 1:1 molar ratio. Determine the moles remaining after reaction (ammonia is in excess)

ammonia ---> 0.00900 mol − 0.00180 mol = 0.00720 mol

3) The ammonia that reacts with the HBr produces ammonium ion. From the chemical equation:

NH3 + HBr ---> NH4Br

We can see that the ammonium ion is produced in a 1:1 molar ratio with each reactant. Since HBr is the limiting reagent, we determine that 0.00180 mole of ammonium ion will be produced. (We may ignore the bromide. It plays no role in the pH.)

4) Use the Henderson-Hasselbalch equation to determine the pH (Where did the 9.248 come from?):

pH = 9.248 + log (0.00720 / 0.00180)

pH = 9.248 + 0.602

pH = 9.850

5) I used the pKa of the ammonium ion (that's the 9.248) in the Henderson-Hasselbalch equation just above. Here is how I determined the value:

KaKb = Kw

(Ka) (1.77 x 10¯5) = 1.00 x 10¯14

Ka = 5.64972 x 10¯10

pKa = −log 5.64972 x 10¯10 = 9.248

Example #8: Determine the pH of a solution prepared by dissolving 0.35 mole of ammonium chloride in 1.0 L of 0.25 M aqueous ammonia. Kb for ammonia equals 1.77 x 10-5

Solution:

1) This is a buffer solution, with a weak base (the ammonia) and the salt of the weak base (the ammonium chloride) in solution at the same time. We will use the Henderson-Hasselbalch Equation to solve this problem.

pH = pKa + log [base / acid]

2) We know the two concentrations:

pH = pKa + log [0.25 / 0.35]

The molarity of the acid (the ammonium chloride is the acid) was arrived at by dividing 0.35 mol by 1.0 L.

3) Notice that a pKa is involved in the Henderson-Hasselbalch Equation, but we have been provided a Kb. In general we need to be very careful in buffer problems to understand which equilibrium constant is provided and which one is needed. Sometimes, we are not given the one we want. The Kb given is for ammonia (NH3). The buffer in this example is made up of NH4Cl and NH3. Therefore, the "acid" in the buffer is the ammonium ion (NH4+), which comes from the ammonium chloride. In other words, we need the Ka (and then the pKa) for the ammonium ion. Here is the procedure to find those:

Kw = KaKb

1.00 x 10¯14 = (Ka) (1.77 x 10¯5)

Ka = 5.65 x 10¯10

pKa = −log 5.65 x 10¯10 = 9.248

By the way, providing one constant (the Kb in this example) while the solution requires a different, but related, constant (the pKa in this example) is very common in problems of this type.

4) Now, we can finish:

pH = 9.248 + log [0.25 / 0.35]

pH = 9.248 + (−0.146) = 9.10

Comment: Kw = KaKb is an important equation to know. It is a favorite trick of teachers to ask you something that requires either the Ka or the Kb, but only give you the other value. They want to know if you know to use Kw = KaKb to get the value you need.

The same thing was done in example #5, except there I used pKa + pKb = pKw.

Example #9: You have 0.500 L of an acetic acid buffer (0.800 M total) at maximum buffering capacity. To it, you add 0.100 mole of HCl. What is the new pH?

Solution:

Note: since the problem is silent about volume change upon addition of the HCl, we assume no volume change.

1) The problem statement describes the original solution as this:

"0.500 L of an acetic acid buffer (0.800 M total) at maximum buffering capacity"

There is a lot of information in that sentence. First, since it is an "acetic acid buffer" we know that both acetic acid CH3COOH (or HAc) and its conjugate base acetate ion CH3COO¯ (or Ac¯) are present. Maximum buffering capacity means that the acid and its conjugate base are in a 1:1 molar ratio.

2) The original composition of the buffer:

0.800 M (total) means 0.800 mol (total) per liter

We have 0.500 L present, which means 0.400 mol (total) present

Maximum buffering capacity requires that HAc and Ac¯ be present in a 1:1 molar ratio. Since 0.400 mol (total) is present, that means this is present in 0.500 L:

moles HAc = 0.200 mol
moles Ac¯ = 0.200 mol

2) The added HCl (being an acid) will react with the base (the acetate). It will do so in a 1:1 molar ratio. New amounts:

moles HAc ---> 0.200 + 0.100 = 0.300 mol
moles acetate ---> 0.200 − 0.100 = 0.100 mol

3) Use the Henderson-Hasselbalch:

pH = 4.752 + log (0.100 / 0.300)

pH = 4.752 + (−0.477)

pH = 4.275

Example #10: You have 0.500 liter of an acetic acid buffer (0.800 M total) at maximum buffering capacity. To it, you add 0.100 mole of solid salt (with no volume change). What is the new pH?

Discussion:

This one is a bit sneaky because the added "salt" is not identified. The word "salt" could be almost anything. Does "salt" mean sodium chloride NaCl? or ammonium chloride NH4Cl? Or potassium formate HCOOK? Or ammonium acetate CH3COONH4? etc.

The most logical salt should be assumed. Here the word "salt" most likely means the "salt of the given weak acid." The weak acid is acetic acid CH3COOH so its salt would contain CH3COO¯. We will assume sodium acetate CH3COONa is added (although potassium acetate and others would work too).

REALLY IMPORTANT: If we had assumed that the word "salt" meant something like NaCl or KBr, then this because a fairly trivial problem. Right? Do you know why it would be trivial? (Answered at the end of the solution.)

Solution:

1) Starting amounts:

HAc ---> 0.400 mol
salt ---> 0.400 mol

2) Ending amounts:

HAc ---> 0.400 mol
salt ---> 0.400 + 0.100 = 0.500 mol

3) Use the Henderson-Hasselbalch:

pH = 4.752 + log (0.500 / 0.400)

pH = 4.752 + 0.097

pH = 4.849

The buffer's pH increases (becoming more basic) because we added a (weak) base to the solution. In Example #9, the opposite occurred because we added a (strong) acid. Notice that I did not bother to change moles to molarities. This step is not required since the volume is the same for the acid in solution as well as the base.

If we added a salt like NaCl or KBr (salts of a strong acid and a strong base), the solution to the problem becomes this:

pH = pKa

This is because salts like NaCl have no effect on the pH of a buffer. They are neutral salts, forming solutions of pH = 7 when dissolved on their own.

Example #11: 4.92 g of a monoprotic weak acid (use HA for its formula) was dissolved in 500. mL of solution and titrated against a 0.500 M solution of NaOH. After 16.0 mL of NaOH solution was added, the pH was observed to be 4.250. The equivalence point was reached after 80.0 mL of base solution had been added.

(a) Calculate the molecular weight of the acid.
(b) Calculate the Ka of the acid.
(c) Calculate the pOH of the original solution.
(d) Calculate the pH after 40.0 mL of NaOH solution was added.
(e) Calculate the percent ionization of the acid in the original solution.
(f) Calculate the pH at the equivalence point.

Note: part (e) is not often asked in the context of a multi-part buffer question.

Comment:

Note that as a result of the titration TWO pieces of data emerge:
(i) After we add 16 mL of the NaOH, the pH = 4.250.
(ii) After we add 80 mL of the NaOH, the equivalence point is reached.

The second piece of data (80 mL) is for part (a) and the first piece of data (16 mL) is for the answer to part (b).

Solution to (a):

1) A monoprotic acid will react with sodium hydroxide in a 1:1 molar ratio:

HA + OH¯ ---> H2O + A¯

This is commonly called a neutralization reaction.

2) The molecular weight requires the moles represented by the 4.92 g of HA. We know 80.0 mL of 0.500 m NaOH was needed to reach the equivalence point. How many moles of NaOH is this?

moles = (0.500 mol/L) (0.0800 L) = 0.0400 mol

3) But what is the equivalence point? This happens during a titration and can be described several ways:

(i) the moles of strong base added (here NaOH) equals the original number of moles of acid (here HA)
(ii) the titrant (here NaOH) is "chemically equivalent" to the analyte (here HA)
(iii) enough NaOH has been added to "neutralize" the original HA present

Based on the 1:1 molar ratio between HA and NaOH (see chemical equation above), we conclude that 0.0400 mol of acid was originally present. In other words, at the equivalence point, the moles of NaOH added equals the moles of HA present before reaction.

4) Determine molecular weight of weak acid:

4.92 g / 0.0400 mol = 123 g/mol

Solution to (b):

We will use the Henderson-Hasselbalch equation to determine the pKa of the acid. From the pKa to the Ka is one additional step.

1) Determine the moles of NaOH added by the addition of 16.0 mL:

moles = (0.500 mol/L) (0.0160 L) = 0.00800 mol

2) This amount of NaOH reacts with the HA, lowering the amount of HA in solution:

0.0400 mol − 0.00800 mol = 0.0320 mol of HA remaining.

Please note that 0.0080 mol of A¯ is produced.

3) We are now ready to use the Henderson-Hasselbalch Equation:

 [base] pH = pKa + log ––––– [acid]

 0.0080 4.250 = pKa + log ––––– 0.0320

Note the direct use of moles rather than molarities.

4.250 = pKa + (−0.602)

pKa = 4.852

Ka = 1.406 x 10¯5

Which, to three sig figs, is 1.41 x 10¯5

4) An alternative solution path to (b) was shown to me by a colleague. It uses the Ka expression:

 [H3O+] [A¯] Ka = –––––––––– [HA]

 (5.623 x 10¯5) (0.00800 + 5.623 x 10¯5) Ka = ––––––––––––––––––––––––––––––– 0.0320 − 5.623 x 10¯5

Notice that this set up includes the amount that the HA would go down and the amount that the A¯ would go up when producing the H3O+ in the solution. The Henderson-Hasselbalch Equation ignores amount, making the H-H an approximation.

The [H3O+] came from the pH of 4.250.

 (5.623 x 10¯5) (0.00805623) Ka = –––––––––––––––––––––– 0.03194377

Ka = 1.418 x 10¯5

Which, to three sig figs, is 1.42 x 10¯5

Solution to (c):

The original solution was simply a weak acid in solution and so technically is not considered a buffer. This is because the amount of conjugate base that would be naturally present at equilibrium would be too small to be an effective buffer. To be an buffer we need comparable amounts of both weak acid and weak base. Even though it is not a buffer, we will be able to get the pOH using the Ka and the original concentrations.

HA + H2O ⇌ H3O+ + A¯

 [H3O+] [A¯] Ka = –––––––––– [HA]

 (x) (x) 1.406 x 10¯5 = ––––––––– 0.0800 − x

Note that I used a slightly less-rounded off value for the Ka. Ignoring the 'subtract x' leads to:

x = $\sqrt{\mathrm{\left(1.406 x 10¯5\right) \left(0.0800\right)}}$

x = 0.001060566 M

pH = 2.974

pOH = 11.026

Solution to (d):

The equivalence point was reached at 80.0 mL of NaOH solution. This means that the half-equivalence point was reached with 40.0 mL of NaOH solution.

At the half-equivalence point, we know by definition that exactly half the HA has been neutralized. This means that, at the half-equivalence point, the concentrations of HA and A¯ are equal.

Because [HA] = [A¯], this means that the Henderson-Hasselbalch Equation simplifies to:

pH = pKa

In other words, the log portion of the H-H became the log of 1, which equals 0. So, the log portion drops out, leaving us with the equation just above.

pH = 4.852

Solution to (e):

Percent ionization = the hydrogen ion concentration divided by the original acid concentration times 100

(0.0010606 M / 0.0800 M) * 100 = 1.32575%

to three sig figs, 1.32%

Solution to (f):

1) At the equivalence point, all the weak acid has been converted to its salt, symbolized by A¯. Since salts of weak acids are bases, we will do a Kb calculation and arrive at a basic pH value.

A¯ + H2O ⇌ HA + OH¯

 [HA] [OH¯] Kb = –––––––––– [A¯]

2) We need to calculate the Kb value as well as [A¯]:

KaKb = Kw

(1.406 x 10¯5) (Kb) = 1.00 x 10¯14

Kb = 7.11238 x 10¯10

[A¯] = 0.0800 mol / 0.580 L = 0.137931 M

Note use of the combined volumes (500 mL of weak acid mixed with 80 mL of NaOH solution.

3) Calculate the [OH¯] of the solution:

 (x) (x) 7.11238 x 10¯10 = –––––––– 0.137931

x = 9.90463 x 10¯6 M

4) Calculate the pOH, then the pH, of the solution:

pOH = −log 9.90463 x 10¯6 = 5.004

pH = 8.996

Example #12: You are titrating 21.00 mL of a 0.850 M solution of NH3 (ammonia) with a strong acid. If you add 1.20 mL of a 1.80 M solution of hydrochloric acid, what is the final pH of the resulting solution? The Kb for ammonia is 1.77 x 10¯5.

Solution:

1) We need to know how many moles of HCl were used:

(1.80 mol/L) (0.00120 L) = 0.00216 mol

2) We need to know how many moles of ammonia were present before reaction:

(0.850 mol/L) (0.02100 L) = 0.01785 mol

3) The H+ will react with the ammonia to produce ammonium (NH4+) ion. Determine how much ammonia remains:

0.01785 mol − 0.00216 mol = 0.01569 mol

By the way, 0.00216 mol of ammonium ion was produced by the H+ reacting with the NH3.

4) We are almost ready to use the Henderson-Hasselbalch Equation. The H-H Equation requires a pKa value, which we obtain from the Kb of ammonia:

Ka Kb = Kw

(Ka) (1.77 x 10¯5) = 1.00 x 10¯14

Ka = 5.64972 x 10¯10

pKa = −log 5.64972 x 10¯10 = 9.248

5) Alternate technique to get the pKa:

pKa + pKb = pKw

pKb = −log 1.77 x 10¯5 = 4.752

pKa + 4.752 = 14

pKa = 9.248

6) Unleash the H-H!!!

 [base] pH = pKa + log ––––– [acid]

 0.01569 pH = 9.248 + log ––––––– 0.00216

pH = 9.248 + 0.861 = 10.109

6) One can show (by a Kb calculation) that the pH of the original ammonia solution before adding HCl was 11.589. Adding the acid should result in the resulting solution becoming more acidic and it did, moving to a new pH of 10.109.

Example #13: Calculate the pH of a buffer solution that contains 0.200 M NH3 (Kb = 1.77 x 10¯5) and 0.150 M NH4Cl after 0.0200 mole of NaOH has been added.

Solution:

1) Use the Henderson-Hasselbalch:

 [base] pH = pKa + log ––––– [acid]

 0.200 pH = 9.248 + log ––––– 0.150

pH = 9.248 + 0.125 = 9.373

Note that I used the pKa of ammonium ion.

2) Here's the solution that was attached to the question. It's based on a misread of the question:

First, you need to know the volume of buffer into which that 0.02 mol of NaOH has been added. This is because you need to know the actual moles of NH3 and NH4+ in the original solution, not just their concentration. So, for example, assume you have 500 mL of that original buffer. In that solution you have:
(0.5 L) (0.20 mol/L) = 0.10 mol NH3
(0.5 L) (0.15 mol/L) = 0.075 mol NH4+

When you add 0.02 mol NaOH, you quantitatively neutralize NH4+ to NH3, so that after the addition,

moles NH3 = 0.10 + 0.02 = 0.12 mol NH3
moles NH4+ = 0.075 − 0.02 = 0.055 mol NH4+

Then, pH = pKa + log (moles NH3 / moles NH4+)

pH = 9.25 + log (0.12 / 0.055) = 9.59

3) The misread comes from the fact that the two molarities (0.200 and 0.150) resulted AFTER the 0.0200 mol of NaOH was added. Tricky!

4) I'd like to continue with the misread discussed in step 2. What if we assume 1.00 L rather than 0.5 L is present? What pH results after 0.02 mol of NaOH is added?

(1.00 L) (0.20 mol/L) = 0.20 mol NH3
(1.00 L) (0.15 mol/L) = 0.15 mol NH4+

moles NH3 = 0.20 mol + 0.02 mol = 0.22 mol
moles NH4+ = 0.15 mol − 0.02 mol = 0.13 mol

pH = 9.25 + log (0.22 / 0.13) = 9.48

5) Like I said: Tricky!

Example #14: A buffer was prepared by adding 7.45 g of NH4Cl to 60.00 mL of 2.32 M NH3 in a 250-mL volumetric flask and diluting the mark with water. The Kb of NH3 is 1.77 x 10¯5. Calculate the pH of the resulting buffer solution.

Solution:

1) We need the molarity of the ammonium chloride:

MV = grams / molar mass

(M) (0.250 L) = 7.45 g / 53.4916 g/mol

M = 0.557097 mol/L

2) We need the molarity of the ammonia:

M1V1 = M2V2

(2.32 mol/L) (60.00 mL) = (x) (250.0 mL)

x = 0.5568 M

3) We are ready for the Henderson-Hasselbalch:

 0.557097 pH = 9.248 + log ––––––– 0.5568

pH = 9.248 + 0.0002316

pH = 9.248

Example #15: You are given an aqueous buffer whose volume is 2.50 L. It contains 0.250 mole of NH3 and 0.225 mole of NH4Cl. What is the pH that is created when a 500. mL solution of ____ M HCl is added to the entire buffer solution? The pKb of NH3 is 4.752.

Comment:

When you add HCl to a buffer, there are three possible outcomes:

(a) The HCl protonates some, but not all, of the NH3. You still have a buffer in that case. In this situation, the Henderson-Hasselbalch Equation is used.

(b) There is exactly enough HCl to neutralize all of the NH3, leaving only NH4Cl in solution. That is a solution of a salt of a weak base, it is not a buffer. Do a Ka caculation using the Ka of NH4+.

(c) There is excess HCl left after all the NH3 has been protonated. In that case, you ignore all the NH4Cl that is in solution and treat the solution as having only a strong acid in it. The solution is not a buffer, nor is it a weak acid calculation. You are simply calculating the pH of a solution of a strong acid.

Below, I will specify a concentration of HCl that will do each of the outcomes.

By the way, three scenarios like above can be developed using NaOH. You will see those below the solution to (c). They are marked (d), (e), and (f).

Solution to (a):

1) I will use 0.150 M HCl for (a). What we do first is determine how many moles of HCl were added:

moles = MV = (0.150 mol/L) (0.500 L) = 0.0750 mol

2) Since the initial moles of NH3 and NH4Cl are given, I will simple list them:

NH3 ---> 0.250 mol
NH4Cl ---> 0.225 mol

3) The HCl reacts with the NH3 to form NH4Cl. We need to determine how much of each is present after the HCl is used up:

NH3 ---> 0.250 mol − 0.0750 mol = 0.175 mol
NH4Cl ---> 0.225 mol + 0.0750 mol = 0.300 mol

Note that HCl reacts with NH3 to form NH4Cl in a 1:1:1 molar ratio.

4) We are now ready for the Henderson-Hasselbalch Equation:

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

pH = 9.248 + log (0.175 / 0.300) <--- note use of moles, no need for molarity because moles are in the same ratio as molarity

pH = 9.248 + (−0.234)

pH = 9.014

Note that I used the pKa of ammonium, not the pKb of ammonia.

5) We can compare 9.014 to the pH of the buffer before any HCl was added:

pH = 9.248 + log (0.250 / 0.225)

pH = 9.248 + 0.046

pH = 9.294

Adding the HCl (an acid) has changed the pH of the buffer in the acidic direction, from 9.294 to 9.014.

Solution to (b):

1) I will use 0.500 M HCl for (b). What we do first is determine how many moles of HCl were added:

moles = MV = (0.500 mol/L) (0.500 L) = 0.250 mol

2) Allow the HCl to react with the NH3, to produce NH4Cl:

NH3 ---> 0.250 mol − 0.250 mol = 0 mol
NH4Cl ---> 0.225 mol + 0.250 mol = 0.475 mol

3) The ammonium ion is an acid . . . :

NH4+ + H2O ⇌ H3O+ + NH3

4) . . . for which we can write a Ka expression:

 [NH3] [H3O+] Ka = –––––––––––– [NH4+]

5) Substituting and solving:

 (x) (x) 5.64937 x 10¯10 = –––––––– 0.158333

x = 0.0000094577 M

pH = −log 0.0000094577 = 5.024

0.158333 M comes from 0.475 mol divided by 3.00 L

Solution to (c):

1) I will use 0.650 M HCl for (c). What we do first is determine how many moles of HCl were added:

moles = MV = (0.650 mol/L) (0.500 L) = 0.325 mol

2) Allow the HCl to react with the NH3, to produce NH4Cl:

NH3 ---> 0.250 mol − 0.325 mol = −0.075 mol

The NH3 is completely consumed and there is 0.075 mol of HCl left over.

3) Determine the molarity of the HCl:

0.075 mol / 3.00 L = 0.025 M

4) Determine the pH:

pH = −log 0.025 = 1.60

Remember, in the presence of a strong acid (HCl), the weak acid (NH4Cl) plays no role in determining the pH.

Here's the problem with NaOH replacing HCl for parts (d), (e), and (f)

You are given an aqueous buffer whose volume is 2.50 L. It contains 0.250 mole of NH3 and 0.225 mole of NH4Cl. What is the pH that is created when a 500. mL solution of ____ M NaOH is added to the entire buffer solution? The pKb of NH3 is 4.752.

Comment:

When you add NaOH to a buffer, there are three possible outcomes:

(a) The NaOH deprotonates some, but not all, of the NH4Cl. You still have a buffer in that case. In this situation, the Henderson-Hasselbalch Equation is used.

(b) There is exactly enough NaOH to neutralize all of the NH4Cl, leaving only NH3 in solution. That is a solution of a weak base, it is not a buffer. Do a Kb caculation using the Kb of NH3.

(c) There is excess NaOH left after all the NH3 has been deprotonated. In that case, you ignore all the NH3 that is in solution and treat the solution as having only a strong base in it. The solution is not a buffer, nor is it a weak acid calculation. You are simply calculating the pH of a solution of a strong base.

Below, I will specify a concentration of NaOH that will do each of the outcomes.

Solution to (d):

1) I will use 0.150 M NaOH for (d). What we do first is determine how many moles of NaOH were added:

moles = MV = (0.150 mol/L) (0.500 L) = 0.0750 mol

2) Since the initial moles of NH3 and NH4Cl are given, I will simple list them:

NH3 ---> 0.250 mol
NH4Cl ---> 0.225 mol

3) The NaOH reacts with the NH4Cl to form NH3. We need to determine how much of each is present after the NaOH is used up:

NH3 ---> 0.250 mol + 0.0750 mol = 0.325 mol
NH4Cl ---> 0.225 mol − 0.0750 mol = 0.150 mol

Note that NaOH reacts with NH4Cl to form NH3 in a 1:1:1 molar ratio.

4) We are now ready for the Henderson-Hasselbalch Equation:

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

pH = 9.248 + log (0.325 / 0.150) <--- note use of moles, no need for molarity here

pH = 9.248 + (0.336)

pH = 9.584

Note that I used the pKa of the ammonium ion, not the pKb of ammonia.

5) We can compare 9.584 to the pH of the buffer before any NaOH was added:

pH = 9.248 + log (0.250 / 0.225)

pH = 9.248 + 0.046

pH = 9.294

Adding the NaOH (a base) has changed the pH of the buffer in the basic direction, from 9.294 to 9.584.

Solution to (e):

1) I will use 0.450 M NaOH for (e). What we do first is determine how many moles of NaOH were added:

moles = MV = (0.450 mol/L) (0.500 L) = 0.225 mol

2) Allow the NaOH to react with the NH4Cl, to produce NH3:

NH3 ---> 0.250 mol + 0.225 mol = 0.475 mol
NH4Cl ---> 0.225 mol − 0.225 mol = 0 mol

3) The ammonia is a base . . .

NH3 + H2O ⇌ NH4+ + OH¯

4) . . . for which we can write a Kb expression:

 [NH4+] [OH¯] Kb = –––––––––––– [NH3]

5) Substituting and solving:

 (x) (x) 1.77 x 10¯5 = –––––––– 0.1583333

x = 0.00167407 M

pOH = −log 0.00167407 = 2.776

pH = 14 − 2.776 = 11.224

0.1583333 M comes from 0.475 mol divided by 3.00 L

Solution to (f):

1) I will use 0.650 M NaOH for (f). What we do first is determine how many moles of NaOH were added:

moles = MV = (0.650 mol/L) (0.500 L) = 0.350 mol

2) Allow the NaOH to react with the NH4Cl to produce NH3:

NH4Cl ---> 0.225 mol − 0.350 mol = −0.125 mol

There is 0.125 mol of NaOH left over.

3) Determine the molarity of the NaOH:

0.125 mol / 3.00 L = 0.04166667 M

4) Determine the pOH:

pOH = −log 0.04166667 = 1.380

pH = 14 − 1.380 = 12.620

Remember, in the presence of a strong base (NaOH), the weak base (NH3) plays no role in determining the pH.

Bonus Problem: You need to prepare an acetate buffer of pH 6.420 from a 0.664 M acetic acid solution and a 2.50 M KOH solution. If you have 975 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.42? The pKa of acetic acid is 4.752.

A colleague has written a Deep Dive into the bonus example solved just below. It is very detailed, delving into several ideas that I have left unexamined below. I recommend you be comfortable with buffer calculations before taking a look.

Solution:

1) Moles of acetic acid:

(0.664 mol/L) (0.975 L) = 0.6474 mol

2) Henderson-Hasselbalch Equation:

pH = pKa + log [base / acid]

6.420 = 4.752 + log [x / (0.6474 − x)]

The x is the moles of acetate that must be present and the 0.6474 − x is the amount of acetic acid.

3) Algebra!

log [x / (0.6474 − x)] = 1.668

[x / (0.6474 − x)] = 46.5586

x = 30.142 − 46.5586x

47.5586x = 30.142

x = 0.63379 mol

Acetate and KOH are in a 1:1 stoichiometric ratio, so this is the required number of moles of KOH.

4) Volume of KOH needed:

0.63379 mol / 2.50 mol/L = 0.253516 L

254 mL seems like a reasonable answer

5) We can try our calculated values and see what happens:

pH = 4.752 + log (0.63379 / 0.01361) = 6.420