Problems #11 - 20

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

(a) CH_{3}COOH & CH_{3}COOK

(b) HClO_{2}& KClO_{2}

(c) NH_{3}& NH_{4}Cl

(d) HClO & KClO

Part B: For the best system, calculate the ratio of

**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 x 90.55 g/mol) = 41.6186 g

denominator (HClO) = (1 mol x 52.4599 g/mol) = 52.46 gThe 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 x 98.1417 g/mol) = 27532.67 g

denominator (HAc) = (1 mol x 60.0516 g/mol) = 60.05 gThe 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) Let use set the pKa 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. mL of 0.432 M methylamine, CH_{3}NH_{2}, is added to 15 mL of 0.234 M HCl?
Assume that the volumes of the solutions are additive. K_{a} = 2.70 × 10^{-11} for CH_{3}NH_{3}^{+}.

**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, CH_{3}NH_{3}Cl).

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) (0.015 L) = 0.00351 mol

CH_{3}NH_{2}---> (0.432 mol/L) (0.040 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 a relatively trivial matter.) 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)The pK

_{a}of CH_{3}NH_{3}^{+}= −log 2.70 x 10^{-11}= 10.5686pH = 10.5686 + log (0.01377 / 0.00351) = 11.1622

11.162 to three sig figs.

By the way, the pH of 0.432 M methylamine is 12.102, so adding some acid (the HCl) to the solution does make the solution more acidic (a pH of 11.162 as compared to 12.102)

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 NaC_{3}H_{5}O_{2}. 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 NaC_{3}H_{5}O_{2}). 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:NaC_{3}H_{5}O_{2}---> (0.10 mol/L) (1.0 L) = 0.10 mol

HCl ---> 0.02 mol <--- it's the limiting reagentAfter the HCl reacts

NaC_{3}H_{5}O_{2}---> 0.10 mol minus 0.02 mol = 0.08 mol

HC_{3}H_{5}O_{2}---> 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 mol0.403 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 mol0.312 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) = 4.03645393

x = 0.0302734 − 4.03645393x

5.03645393x = 0.00302734

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 HNO_{2} 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 HNO_{2} to react with the NaOH to form some NaNO_{2}:

moles HNO_{2}remaining ---> 0.529 − 0.246 = 0.283 mol

moles NaNO_{2}formed ---> 0.246 molHNO

_{2}and NaOH react in a 1:1 molar ratio and form NaNO_{2}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 HNO_{2}, how many grams of NaNO_{2} would you have to add to the solution for a pH of 3.32?

**Solution:**

Note: we will assume that the addition of solid NaNO_{2} 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 NaNO_{2} 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 3.5 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 = pK _{a}+ 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. Substituting the previous equation into this one gives:

1.346[HA] + [HA] = 0.335[HA] = 0.143 M and by subtraction, [A¯] = 0.192 M

3) 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

Ten Buffer Examples | Buffer Problems 1-10 | Buffer Problems 21-30 | Buffer Problems 31-40 |

Intro. to the Henderson-Hasselbalch Equation | Return to the Acid Base menu |