Problems #21 - 30

Fifteen Buffer Examples | Buffer Problems 1-10 | Buffer Problems 11-20 | Buffer Problems 31-40 |

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

**Problem #21:** What volume of 6.00 M NaOH must be added to 0.250 L of 0.300 M HNO_{2} to prepare a pH = 4.00 buffer?

**Solution:**

1) The first thing to do is look up the K_{a} for nitrous acid, to find:

4.0 x 10¯^{4}Several different values can be found. I selected the one above since it seemed more common than the others.

2) Some comments on the chemistry involved:

We know that this reaction will take place:HNO_{2}+ OH¯ ---> H_{2}O + NO_{2}¯The hydroxide MUST be the limiting reagent. Why? If some NaOH was left over, the solution would be a mixture of a strong base (the NaOH) and some NaNO

_{2}(the salt of a weak acid). This is NOT a buffer and the pH would be calculated using the concept of a strong base.So the final solution will be a mixture of HNO

_{2}and NO_{2}¯. Since this solution is a buffer, the Henderson-Hasselbalch equation will be employed.

3) Filling in the H-H, step 1:

pH = pK_{a}+ log [base / acid]We know we must have a buffer of pH = 4.00:

4.00 = pK

_{a}+ log [base / acid]

4) Filling in the H-H, step 2:

4.00 = 3.40 + log [base / acid]3.40 is the pK

_{a}for HNO_{2}.

5) Filling in the H-H, step 3:

I'm going to give you the log portion and then comment on it:

x 4.00 = 3.40 + log –––––––––– 0.0750 − x The unknown 'x' is the amount of NO

_{2}¯ that got produced by the HNO_{2}reacting with the OH¯.The 0.0750 part of (0.0750 − x) comes from this calculation:

(0.300 mol/L) (0.250 L) = 0.0750 moland the 'x' is the moles of HNO

_{2}that reacted. The amount of HNO_{2}lost is equal to the amount of NO_{2}¯ gained.

6) Algebra!

4.00 = 3.40 + log [x / (0.0750 − x)]log [x / (0.0750 − x)] = 0.60

x / (0.0750 − x) = 3.98

x = 0.2985 − 3.98x

4.98x = 0.2985

x = 0.05994 moles <--- don't round off yet

By the 1:1 molar ratio of the overall reaction, that's the moles of hydroxide that need to be added.

7) Determine the volume of sodium hydroxide solution required:

0.05994 mol / 6.00 mol/L = 0.00999 LBased on the K

_{a}value, 2 sig figs seems best ---> 10. mL

**Problem #22:** If an acetate buffer solution was going to be prepared by neutralizing HC_{2}H_{3}O_{2} with 0.10 M NaOH, what volume (in mL) of 0.10 M NaOH would need to be added to 10.0 mL of 0.10 M HC_{2}H_{3}O_{2} to prepare a solution with pH = 5.50?

**Solution:**

Comment: In doing the salt (sodium acetate) and the acid (acetic acid), I'm going to use moles rather than molarity. Since everything occurs in the same volume of solution, the ratio of salt moles to acid moles is the same as the ratio of molarities. Besides, we don't know the final molarities since we are adding an unknown volume of NaOH solution.

1) We need to know the initial moles of acetic acid in the solution:

(0.010 L) (0.1 mol/L) = 0.001 mol

2) Let us insert values into the H-H equation:

pH = pK_{a}+ log [base / acid]5.50 = 4.752 + log [x / (0.001 − x)]

4.752 is the pK

_{a}of acetic acid

x is the moles of sodium acetate produced by the NaOH reacting

0.001 − x is the amount of acetic acid remaining in solution.The moles of acetate will give us moles of NaOH since there is a 1:1 molar ratio between the two.

3) Continue solving:

log (x / 0.001 − x) = 0.748

4) Antilog both sides

(x / 0.001 − x) = 5.598

5) Cross multiply & simplify to get:

6.598x = 5.598 x 10¯^{3}x = 8.5 x 10¯

^{4}moles

6) Let us determine the volume of NaOH required:

8.5 x 10¯^{4}mol divided by 0.1 mol/L = 8.5 x 10¯^{3}L = 8.5 mL

**Problem #23:** A beaker with 175 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.40 mL of a 0.300 M HCl solution to the beaker. What is the new pH? The pK_{a} of acetic acid is 4.752.

**Solution:**

1) Use the Henderson-Hasselbalch to get the molarities of the base and the acid:

pH = pK_{a}+ log [base / acid]5.000 = 4.752 + log [x / (0.1 − x)]

Note that x and 0.1 − x add up to 0.1, which is the total molarity

log [x / (0.1 − x)] = 0.248

[x / (0.1 − x)] = 1.7701

x = 0.17701 − 1.7701x

2.7701x = 0.17701

x = 0.0639 M <--- that's the acetate concentration in the pH 5 buffer

0.1 − 0.0639 = 0.0361 <--- that's the acetic acid concentration

2) Now, we need to know the moles of acetic acid, acetate and HCl:

acetate ---> (0.0639 mol/L) (0.175 L) = 0.0111825 mol

acetic acid ---> (0.0361 mol/L) (0.175 L) = 0.0063175 mol

HCl ---> (0.300 mol/L) (0.00840 L) = 0.00252 mol

3) The HCl will react with the acetate and turn it into acetic acid. The acetate amount goes down and the acetic acid goes up. Note that all the reaction stoichiometries are 1:1.

acetate ---> 0.0111825 mol − 0.00252 mol = 0.0086625 mol

acetic aid ---> 0.0063175 mol + 0.00252 mol = 0.0088375 mol

4) Now, for the HH equation again:

pH = 4.752 + log (0.0086625 / 0.0088375)pH = 4.752 + log 0.98019802

pH = 4.752 + (−0.009)

pH = 4.743

**Problem #24:** 200.0 mL of an acetate/acetic acid buffer is 0.100 M in total molarity and has a pH of 5.000. After 6.30 mL of 0.490 M HCl is added, what is the new pH?

**Solution:**

1) We ned to know the amount of acetic acid (HAc) and acetate ion (Ac¯) in the pH 5 buffer:

pH = pK_{a}+ log [base / acid]5.000 = 4.752 + log [x / (0.02 − x)]

0.02 comes from:

MV = (0.100 mol/L) (0.2000 L) = 0.0200 mollog [x / (0.02 − x)] = 0.248

x / (0.02 − x) = 1.770109

x = 0.03540218 − 1.770109x

2.770109x = 0.03540218

x = 0.01278 mol <--- that's the acetate moles

0.02 − 0.01278 = 0.00722 mol <--- that's the acetic acid moles

2) Determine the moles of HCl to be added:

MV = (0.490 mol/L) (0.00630 L) = 0.003087 mol

3) The HCl will protonate the Ac¯, causing its amount to decrease and the HAc amount to increase.

Ac¯ ---> 0.01278 − 0.003087 = 0.009693 mol

HAc ---> 0.00722 + 0.003087 = 0.010307 mol

4) Use the Henderson-Hasselbalch Equation to calculate the new pH:

pH = pK_{a}+ log [base / acid]pH = 4.752 + log [0.009693 / 0.010307]

pH = 4.752 + [−0.027]

pH = 4.725

**Problem #25a:** We desire to make a pH 5.000 buffer and we choose a weak acid (let's call it HA) with a pK_{a} of 4.700. Starting with 0.100 M each HA and NaA, we desire to make 100. mL buffer solution.

**Solution:**

1) Use the Henderson-Hasselbalch equation:

5.000 = 4.700 + log [A¯] / [HA][A¯] / [HA] = 10

^{0.300}[A¯] / [HA] = 2.00

2) Use the definition of molarity:

M = moles / volumemoles = MV

moles of A¯ = (0.100 mol / L) (L

_{A¯})

moles of HA = (0.100 mol / L) (L_{HA})

3) Our ratio now becomes:

[(0.100 mol/L) (L_{A¯})] / [(0.100 mol/L) (L_{HA})] = 2.00(L

_{A¯}) / (L_{HA}) = 2.00

4) Set variables and substitute into above ratio:

let L_{A¯}= x

therefore L_{HA}= 0.1 − xComment: I used 0.1 because total volume = 100. mL or 0.1 L.

x / (0.1 − x) = 2.00

x = 0.2 − 2x

3x = 0.2

x = 0.667 L

We require 66.7 mL of NaA and 33.3 mL of HA to make our pH 5 buffer.

**Problem #25b:** Determine how you would prepare 1.00 L of this buffer starting with 0.100 M HA, 0.100 M NaOH and water where the total concentration of HA plus NaA is 0.010 M.

**Solution:**

pH = pK_{a}+ log [base / acid]The total moles of HA and NaA will be 0.010 mol. This comes from the total molarity (0.0100 M) times the final volume of the solution (1.00 L). I will use moles in the log portion of the Henderson-Hasselbalch Equation.

5.000 = 4.700 + log [x / (0.01 − x)]

0.300 = log [x / (0.01 − x)]

x / (0.01 − x) = 2

x = 0.02 − 2x

x = 0.00667 mole of the base (the A¯)

Comments:

(a) x represents the moles of the NaA (the salt) in the 1.00 L of solution

(b) 0.010 − x = 0.00333; this is the moles of HA in 1.0 L of solution.How to prepare the buffer:

(i) Take 100. mL of 0.100 M HA. This represents 0.0100 mole of HA. (ii) Add 66.7 mL of 0.100 M NaOH solution. This is 0.0067 mole of NaOH. The NaOH reacts with the HA to form NaA, the salt of HA. (iii) Dilute to 1.00 L with water.

**Problem #26:** Calculate the volume (in mL) of 0.170 M NaOH that must be added to 311 mL of 0.0485 M HA (a generic weak acid) to give the solution a pH of 7.55. The pK_{a} of HA = 7.18.

**Solution:**

1) Since the moles of HA and the salt formed from HA/NaOH (for which I will use A¯) are in the same volume of solution, we can use moles in the log portion of the Henderson-Hasselbalch equation:

7.55 = 7.18 + log (A¯) / (HA − A¯)

2) The moles of A¯ will be the unknown:

moles HA ---> (0.0485 mol/L) (0.311 L) = 0.0150835 mol

3) I'll use the unrounded off number.

7.55 = 7.18 + log [(x) / (0.0150835 − x)]0.37 = log (x) / (0.0150835 − x)

4) antilog both sides

(x) / (0.0150835 − x) = 2.34423

5) cross multiply

x = 0.035359 − 2.34423x3.34423x = 0.035359

x = 0.0105732 moles of A¯ required

6) mL of NaOH required:

0.0105732 mol / 0.170 mol/L = 0.0621953 L = 62.2 mL (to three sig figs)

**Problem #27:** What mass of HCl would need to be added to a 250. mL solution containing 0.500 M NaC_{2}H_{3}O_{2} and 0.500 M HC_{2}H_{3}O_{2}, to make the pH = 4.25? K_{a} of HC_{2}H_{3}O_{2} is 1.77 x 10^{-5}.

**Solution with moles:**

pH = pK_{a}+ log [base / acid]4.25 = 4.752 + log [base / acid]

−0.502 = log [base / acid]

[base/acid] = 0.314775 <--- this is the base/acid ratio needed to create a pH of 4.25

(0.125 − x) / (0.125 + x) = 0.314775 <--- note use of total moles in solution as opposed to molarity

0.039346875 + 0.314775x = 0.125 − x

1.314775x = 0.085653125

x = 0.0651466 mol

(0.0651466 mol) (36.5 g/mol) = 2.38 g

**Solution with molarities:**

(0.5 − x) / (0.5 + x) = 0.3147750.1573875 + 0.314775x = 0.5 − x

1.314775x = 0.3426125

x = 0.2605864 M <--- the molarity of the HCl that must be achieved to create the desired pH of 4.25

MV = g/molar mass

(0.2605864) (0.25) = x / 36.5

x = 2.38 g

**Problem #28:** What mass of HCl would need to be added to a 250. mL solution containing 0.500 M NaC_{2}H_{3}O_{2} and 0.500 M HC_{2}H_{3}O_{2}, to make the pH = 4.25? K_{a} of HC_{2}H_{3}O_{2} is 1.77 x 10¯^{5}

**Solution:**

pH = pK_{a}+ log [base / acid]4.25 = 4.752 + log [base / acid]

−0.502 = log [base / acid]

[base / acid] = 0.314775 <--- we need to create a solution with this base/acid ratio to get our pH of 4.25

(0.125 − x) / (0.125 + x) = 0.314775 <--- the 0.125 is the moles of each solute [from (0.5 mol/L) (0.25 L)]

The HCl converts the acetate (the base) into the acid (the acetic acid)

0.04001375 + 0.314775x = 0.125 − x

1.314775x = 0.08498625

x = 0.0646394 mol

(0.0646394 mol) (36.4609 g/mol) = 2.36 g

**Problem #29:** How many mL of 0.75 M HCl must be added to 120 mL of 0.90 M sodium formate to make a buffer of pH = 4.00? pK_{a} of formic acid = 3.75

**Solution:**

1) I'll use the Henderson-Hasselbalch equation to solve the problem, but the amounts of acid and base will be expressed in terms of moles, not molarities.

pH = pK_{a}+ log [base / acid]

2) Determine moles formate:

(0.90 mol/L) (0.12 L) = 0.108 mol <--- formate is the base in this problem.

3) The HCl will protonate some of the formate, making formic acid. So, the formate amount will go down from 0.108 mol and the formic acid amount will go up from zero.

4.00 = 3.75 + log (0.108 − x) / xlog (0.108 − x) / x = 0.25

(0.108 − x) / x = 1.77828

1.77828x = 0.108 − x

2.77828x = 0.108

x = 0.038873 mol <--- this is the amount of formic acid formed

4) Since HCl reacted and formic acid formed are in a 1:1 molar ratio, we can determine the volume of HCl required:

0.038873 mol / 0.75 mol/L = 0.05183 L = 51.8 mL5) As a check, let's insert the moles back into the H-H equation as follows:

pH = 3.75 + log (0.069127 / 0.038873)pH = 3.75 + 0.25 = 4.00

You can also use molarities (by dividing moles by the total volume of 0.1718 L).

**Problem #30a:** You need to prepare a buffer solution of pH 4.178 from 25.0 mL of 0.282 M solution of a sodium salt of a weak acid, NaA where the pK_{a} of the weak acid HA is 4.270. What volume of 0.329 M HCl would you need to add?

**Solution:**

1) The Henderson-Hasselbalch Equation is this:

[base] pH = pK _{a}+ log––––– [acid]

2) We already know two values:

[base] 4.178 = 4.270 + log ––––– [acid]

3) The anion of the weak acid (the A¯) is the base and HA will be the acid. We know that adding some HCl will turn some of the A¯ into HA. Let's determine how much A¯ we have present:

(0.282 mol/L) (0.025 L) = 0.00705 mol

4) That amount of A¯ is going to go down by some unknown amount when it reacts with the HCl. But, here's the key: the HA amount will go up by the exact same amount. This allows me to fill in the H-H equation:

0.00705 − x 4.178 = 4.270 + log –––––––––– x 0.00705 − x ---> that's the amount of A¯ remaining (after all the HCl is used up)

x ---> that's the amount of HA produced

5) This whole thing works because of the 1:1 molar ratio between A¯ used up and HA produced. Now, for some algebra:

0.00705 − x log –––––––––– = −0.092 x

0.00705 − x –––––––––– = 0.809096 x 0.00705 − x = 0.809096x

1.809096x = 0.00705

x = 0.00389697 mol <--- the moles of HA needed in the base/acid ratio

6) Because of the 1:1 molar ratio in the chemical reaction, it's also the moles of HCl we need. On to the volume of HCl:

0.00389697 mol / 0.329 mol/L = 0.0118449 L = 11.8449 mLTo three sig figs, the answer is 11.8 mL

**Problem #30b:** You need to prepare an acetate buffer of pH 5.83 from a 0.642 M acetic acid solution and a 2.31 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 5.830? The pK_{a} of acetic acid is 4.752.

**Solution:**

1) We will use the Henderson-Hasselbalch equation:

pH = pK_{a}+ log [base / acid]

2) Here is the H-H, set up with what we know:

5.830 = 4.752 + log [base / acid]

3) The base and the acid amounts will be expressed in moles. The base, by the way, is the acetate anion, not the KOH.

We do not know how much acetate is required, so we will call it x.The acetic acid amount is this:

0.62595 − xThe 0.62595 mol comes from this:

(0.642 mol/L) (0.975 L) = 0.62595 molThe minus x come from the fact that some of the acetic acid will be converted to acetate in a 1:1 molar ratio.

4) We are now ready for the completed H-H:

5.830 = 4.752 + log [x / (0.62595 − x)]

5) Now, for a bit of algebra:

log [x / (0.62595 − x)] = 1.078x / (0.62595 − x) = 11.9674

x = 7.490994 − 11.9674x

12.9674x = 7.490994

x = 0.57768 mol of acetate required

6) Because of the 1:1 molar ratio between OH¯ consumed and acetate produced, the moles of acetate equals the moles of KOH required.

0.57768 mol divided by 2.31 mol/L = 0.250078 L0.250078 L = 250. mL (to three sig figs)

**Bonus Problem:** pK_{a} for phenophthalein is 9.3 at room temp.

(a) Calculate ratio of its anionic form to acid form at pH 8.2 and at pH 10.

(b) Using these values, explain the colour change within this pH range.

**Solution to part (a):**

1) At pH = 8.2:

8.2 = 9.3 + log (base form / acid form)log (base form / acid form) = −1.1

ratio of base form to acid form = 0.0794 to 1 (call it 8 to 100)

2) At pH = 10.:

10 = 9.3 + log (base form / acid form)log (base form / acid form) = 0.7

ratio of base form to acid form = 5.01 to 1 (call it 500 to 100)

**Solution to part (b):**

1) Key fact:

it's the anionic (or base) form that is colored pink. The acid form is colorless.

2) At pH = 8.3:

the pink form is in the minority at this pH. For every 100 acid (colorless) forms, there are only 8 base (pink) forms.

3) At pH = 10.:

the colorless form is in the minority. For every 100 acid (colorless) forms present, there are now 500 base (pink) forms present.

4) This means:

From pH = 8.3 to pH = 10., there has been a 6250% increase in pink forms (from 8:100 to 500:100).While there might be a slight pink color (viewed against a white background) at pH 8.3, the population of pink forms has greatly increased by pH = 10., to the point where the pink color is now easily seen, even without the aid of a white background.

Fifteen Buffer Examples | Buffer Problems 1-10 | Buffer Problems 11-20 | Buffer Problems 31-40 |

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