### Buffers and the Henderson-Hasselbalch EquationProblems #31 - 40

Problem #31: How much sodium formate (HCOONa, 68.0069 g/mol) do you need to add to 400. mL of 1.00 M formic acid for a pH 3.500 buffer. Ka = 1.77 x 10¯4

Solution:

pH = pKa + log [base/acid]

3.500 = 3.752 + log (x / 1) <--- that 1 comes from the 1.00 M formic acid

log x = −0.252

x = 0.560 M (this is the molarity of the formate that is required)

MV = g/molar mass

(0.560 mol/L) (0.400 L) = x / 68.0069 g/mol

x = 15.2 g

Problem #32: Aspirin has a pKa of 3.4. What is the ratio of A¯ to HA in:

(a) the blood (pH = 7.4)
(b) the stomach (pH = 1.4)

General comment about the solutions: You have to find the ratio between A¯ and HA so the concentrations are not needed

Solution to part (a):

pH = pKa + log ([A¯] / [HA])

7.4 = 3.4 + log ([A¯] / [HA])

7.4 − 3.4 = log ([A¯] / [HA])

4 = log ([A¯] / [HA])

104 = [A¯] / [HA]

10000 A¯ for every one HA

Solution to part (b):

1.4 = 3.4 + log ([A¯] / [HA])

1.4 − 3.4 = log ([A¯] / [HA])

−2.0 = log ([A¯] / [HA])

10¯2 = [A¯] / [HA]

One A¯ for every 100 HA

Problem #33: Calculate the pH of the solution that results from the addition of 0.040 moles of HNO3 to a buffer made by combining 0.500 L of 0.380 M HC3H5O2 (Ka = 1.30 x 10¯5) and 0.500 L of 0.380 M NaC3H5O2. Assume addition of the nitric acid has no effect on volume.

Solution:

1a) The nitric acid will reduce the amount of NaC3H5O2:

(0.380 mol/L) (0.500 L) = 0.190 mol of NaC3H5O2

0.190 mol − 0.040 mol = 0.150 mol NaC3H5O2 remaining

1b) The reaction in 1a will increase the amount of HC3H5O2:

the increase will be by the same amount of the decrease

0.190 mol + 0.040 mol = 0.230 mol of HC3H5O2

2) Calculate the new pH:

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

pH = 4.886 + log (0.150 / 0.230)

pH = 4.700

Note: the ratio of the moles is the same as the ratio of the molarities. You can demonstrate that to yourself by calculating the new molarities in 0.500 L, them adding the two solutions together, thereby cutting the molarities in half. Then, insert them into the Henderson-Hasselbalch Equation. You will get 4.700 for your answer.

Problem #34: You need to produce a buffer solution that has a pH of 5.270. You already have a solution that contains 10.0 mmol (millimoles) of acetic acid. How many millimoles of sodium acetate will you need to add to this solution? The pKa of acetic acid is 4.752.

Solution:

Substitute into the Henderson-Hasselbalch Equation and solve:

5.270 = 4.752 + log (x / 10)

log (x / 10) = 0.518

x / 10 = 3.296

x = 33.0 millimoles of sodium acetate

A colleague has prepared a discussion concerning an exact solution to #34 (as compared to the above approximate solution. You will find it here.

Problem #35: What is the pH when 25.0 mL of 0.200 M of CH3COOH has been titrated with 35.0 mL of 0.100 M NaOH?

Solution:

1) Determine moles of acetic acid and NaOH before mixing:

CH3COOH: (0.200 mol/L) (0.0250 L) = 0.00500 mol
NaOH: (0.100 mol/L) (0.0350 L) = 0.0035 mol

2) Determine moles of acetic acid and sodium acetate after mixing:

CH3COOH: 0.00500 mol − 0.00350 mol = 0.00150 mol
CH3COONa: 0.0035 mol

3) Use the Henderson-Hasselbalch Equation:

pH = 4.752 + log [(0.00350 mol/0.060 L) / (0.0015 mol/0.060 L)]

pH = 4.752 + log 2.333

pH = 4.752 + 0.368 = 5.120

Note the inclusion of the volumes. They simply cancel out, so most HH problems can be solved with moles rather than molarities.

Problem #36: A solution containing 50.0 mL of 0.180 M NH3 (Kb = 1.77 x 10¯5) is being titrated with 0.360 M HCl. Calculate the pH:

(a) initially
(b) after the addition of 5.00 mL of HCl
(c) after the addition of a total volume of 12.5 mL HCl
(d) after the addition of a total volume of 25.0 mL of HCl
(e) after the addition of 26.0 mL of HCl

An example of the solution technique for part (a) is provided here. The answer for our example is 11.252.

Solution to part (b):

1) Determine moles of NH3 and HCl before mixing:

NH3 ---> (0.180 mol/L) (0.0500 L) = 0.00900 mol
HCl ---> (0.360 mol/L) (0.00500 L) = 0.00180 mol

2) Determine moles of ammonia and ammonium ion after mixing:

NH3 ---> 0.00900 mol − 0.00180 mol = 0.00720 mol NH3 remaining
NH4+ ---> 0.00180 mol produced

3) Use the H-H Equation:

pH = pKa + log [base/acid]

pH = 9.248 + log [(0.0072/0.055) / (0.0018/0.055)]

pH = 9.850

Note that the pKa was used, NOT the pKb. Note also that the addition of acid caused the pH of the (now) buffer to decrease, that is to become more acidic. This is what we expect to happen. Lastly, note the inclusion of the new volume, 0.055 L. Both these values cancel out, so the new volume is rather inconsequential. In fact, when I calculated the answer for this problem (March 23, 2010), I actually divided 72 by 18.

Solution to part (c):

1) Determine moles of NH3 and HCl before mixing:

NH3 ---> (0.180 mol/L) (0.0500 L) = 0.00900 mol
HCl ---> (0.360 mol/L) (0.01250 L) = 0.00450 mol

2) Determine moles of ammonia and ammonium ion after mixing:

NH3 ---> 0.00900 mol − 0.00450 mol = 0.00450 mol NH3 remaining
NH4+ ---> 0.00450 mol produced

3) Use the H-H Equation:

pH = pKa + log [base/acid]

pH = 9.248 + log [0.00450 / 0.00450]

pH = 9.248

Solution to part (d):

1) Determine moles of NH3 and HCl before mixing:

NH3 ---> (0.180 mol/L) (0.0500 L) = 0.00900 mol
HCl ---> (0.360 mol/L) (0.02500 L) = 0.00900 mol

2) Determine moles of ammonia and ammonium ion after mixing:

NH3 ---> 0.00900 mol − 0.00900 mol = zero mol NH3 remaining
NH4+ ---> 0.00900 mol produced

3) This solution is no longer a buffer. It is now the salt of a weak base and its solution will be acidic. The H-H Equation is not used to determine the pH of the solution. An introduction to the solution technique can be found here. The actual molarity is important and it is 0.120 M (0.00900 mol / 0.0750 L). The pH is 5.084.

Solution to part (e):

This solution is now a solution of a strong acid (the presence of the weak acid NH4+ may be ignored).

1) Let us determine the new molarity of the HCl:

M1V1 = M2V2

(0.36) (1.00 mL) = (x) (76.00 mL)

x = 0.0047368 M (I kept some guard digits.)

2) Calculate the pH in the usual manner for a strong acid:

pH = −log [H3O+]

pH = 2.324

Problem #37: A beaker with 100.0 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 7.30 ml of a 0.360 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.752.

Solution:

1) Use H-H Equation to determine the ratio of base to acid:

pH = pKa + log (base / acid)

5.000 = 4.752 + log (x / 1)

I want the ratio of base to acid which is why I used x / 1

log x = 0.248

x = 1.77

the base to salt ratio is 1.77:1

2) Determine moles of base and moles of acid in 0.0100 mole of total solute:

let x = moles of acid

1.77x + x = 0.0100 total moles of solute

x = 0.00361 moles (this is the acid)
base = 0.00639 mol

N.B. the 0.0100 total moles resulted from this calculation:

(0.100 mol/L) (0.1000 L) = 0.0100 mol

3) Add strong acid, which will protonate the base:

(0.360 mol/L) (0.00730 L) = 0.002628 mol of strong acid added

base: 0.00639 − 0.002628 = 0.003762 mol
acid: 0.00361 + 0.002628 = 0.006238 mol

pH = 4.752 + log (0.003762/0.006238)

pH = 4.752 + (−0.220)

pH = 4.532 (to three sig figs)

Note: I used moles in the last H-H expression rather than using the new volume of 0.1073 L to get molarities. Since there would be a 0.1073 L in the numerator and the denominator, I just skipped that step.

Problem #38: 1.00 L of a solution containing 0.0500 mole of HAc and 0.100 mole of NaAc is prepared. Ignore the autoionization of water for the purposes of this problem. The Ka of HAc equals 1.77 x 10-5.

(a) Calculate the numerical value of the reaction quotient, Q for the initial condition.
(b) Which way will the reaction shift?
(c) Calculate to 3 significant digits the pH of this solution.

Note: The chemical reaction to consider is:

HAc ⇌ H+ + Ac¯

Solution to (a):

1) Write the reaction quotient for the above reaction and solve for Q:

Qa = ([H+]o [Ac¯]o) / [HAc]o

Qa = [(1.00 x 10¯7) (0.100)] / 0.0500

Qa = 2.00 x 10¯7

Notice that I did use the value for the [H+] derived from the autoionization of water. The instruction in the question is intended for use when the pH is calculated.

Solution to (b):

Since Qa < Ka, the reaction shifts right, in order to increase the value for Qa up to the value for Ka.

Solution to (c):

1) Write the Ka expression and solve it:

Ka = ([H+] [Ac¯]) / [HAc]

1.77 x 10¯5 = [(x) (0.100)] / 0.0500

x = 8.85 x 10¯6 M

2) Calculate the pH:

pH = −log 8.85 x 10¯6 = 5.053

Notice that I did not use the Henderson-Hasselbalch equation. Ah, but I did! The H-H equation is simply a rearrangement of the Ka expression.

Problem #39: Calculate the ratio of CH3NH2 to CH3NH3Cl required to create a buffer with pH = 10.14

Solution:

1) We need the Ka of the methylammonium ion:

Kb of CH3NH2 = 4.4 x 10-4

Ka for CH3NH3Cl = 1.00 x 10-14 / 4.4 x 10-4 = 2.27 x 10-11

2) Write the chemical equation and the Henderson-Hasselbalch equation:

CH3NH3+ + H2O ⇌ H3O+ + CH3NH2

pH = pKa + log (base/acid)

3) Substitute and solve for the base/acid ratio:
10.14 = 10.644 + log (base/acid)

log (base/acid) = −0.504

base/acid = 0.313

Problem #40: Fifty percent of a weak acid is in an ionized form in a solution with pH of 5.000, what is the pKa value for the weak acid?

Solution:

1) I found this answer on-line:

"Let us suppose the acid is HA and its ionic anion is A¯. Then, for a 50% ionized solution, pH = pKa because, at half-neutralization, [HA]/[A¯] = 1. Since log(1) = 0, the pH at half-neutralization is numerically equal to pKa.

Therefore:

pKa = 5.000

Ka = 10¯5.000 = 1.00 x 10¯5

(a) I decided to put the answer among Henderson-Hasselbalch questions because of the reference to the log term of the H-H equation.
(b) The half-neutralization point, where [HA] = [A¯], is an important one in titration and is a technique whereby the pKa of a weak acid can be obtained.
(c) Almost certainly, your test will include a question which utilizes the half-neutralization point.

Bonus Problem: What is the pH at each of the given points in the titration of 50.0 mL of a 0.85 M solution of the generic dibasic base B(aq) with 0.85 M HCl(aq)? (The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.53.) (Note: each volume of HCl added is the total volume added. The volumes are not cumulative.)

(a) before addition of any HCl?
(b) after addition of 25.0 mL HCl?
(c) after addition of 50.0 mL HCl?
(d) after addition of 75.0 mL HCl?
(e) after addition of 100.0 mL HCl?

Solution to (a): a weak base (B) ionizing in solution

1) Write the chemical equation for thefirst step of B ionizing in solution:

B + H2O ⇌ BH+ + OH¯

We do not need to consider the second ionization. This is because the hydroxide ion contribution from the second ionization step is nearly 270,000 times smaller than from the first. We know this from the difference between the two pKb values.

2) Write the Kb1 expression, put values in place, and solve:

Kb1 = ([BH+] [OH¯]) / [B]

0.0079433 = [(x) (x)] / 0.85

x = 0.0821694 M

3) Determine the pOH, then the pH:

pOH = −log 0.0821694 = 1.08

pH = 12.91

Solution to (b): the half-equivalence point for the first Kb

1) Notice that the molarities are the same and that the volume of HCl is exactly half that of the base B. 50% of the base B is protonated to make BH+. When the solution has 50% base (that's B) and 50% conjugate acid (that's BH+), the log portion of the Henderson-Hasselbalch cancels out, leaving this:

pH = pKa

2) Determine the Ka:

Kb = 10¯Kb = 10¯2.10 = 0.0079433

KaKb = Kw

(Ka) (0.0079433) = 1.00 x 10¯14

Ka = 1.259 x 10¯12

3) Determine the pH:

pH = −log 1.259 x 10¯12 = 11.90

Solution to (c): a weak base (BH+) ionizing in solution (note that all B has been converted to BH+)

1) Write the chemical equation for the second step of B ionizing in solution:

BH+ + H2O ⇌ BH22+ + OH¯

2) As the writer of how to solve this problem, I know a Kb calculation is coming up. I also know that it is NOT a buffer, so that means I must know the concentration of the solution. I know the volume doubled, so . . .

M1V1 = M2V2

(0.85 mol/L) (50.0 mL) = (x) (100.0 mL)

x = 0.425 M

2) Write the Kb2 expression, put values in place, and solve:

Kb2 = ([BH22+] [OH¯]) / [BH+]

2.9512 x 10¯8 = [(x) (x)] / 0.425

x = 0.00011199 M

3) pOH, then pH:

pOH = −log 0.00011199 = 3.95

pH = 10.05

Solution to (d): the half-equivalence point for BH+. Half the solution is BH+, half is BH22+

1) We need the pKa for BH22+:

pKa * pKb = pKw

(pKa) (7.53) = 14.00

pKa = 6.47

2) Determine the pH:

pH = pKa

pH = 6.47

Solution to (e): after all the BH+ is converted to BH22+ we then have a weak acid (BH22+) ionizing in solution

1) Determine the molarity of the BH22+:

M1V1 = M2V2

(0.85 mol/L) (50.0 mL) = (x) (150.0 mL)

x = 0.28333 M

2) Write the chemical equation for BH2+ ionizing in solution:

BH22+ + H2O ⇌ BH+ + H3O+

3) Write the Ka expression, put values in place, and solve for the pH:

Ka = ([BH+] [H3O+]) / [BH22+]

3.38845 x 10¯7 = [(x) (x)] / 0.28333

x = 0.00030985 M

pH = −log 0.00030985

pH = 3.51