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Problem #1: 20.0 cm3 of 1.00 M HNO2 and 40.0 cm3 of 0.500 M NaNO2 are mixed. What is the pH of the resulting solution? pKa of nitrous acid is 3.34
Comment: this is an answer that does not mention the Henderson-Hasselbalch Equation. In one's travels, one occasionally runs across an individual that does not like the H-H and insists on using the Ka expression. Just be aware that the Ka expression and the Henderson-Hasselbalch Equation are the same thing.
Solution:
1) This is the chemical reaction of interest:
HNO2 + H2O ⇌ H3O+ + NO2¯
2) The HNO2 is diluted by the addition of NaNO2 solution.
M1V1 = M2V2(1.00 mol/L) (20.0 cm3) = (x) (60.0 cm3) = 0.333 M HNO2 (the initial concentration of the HNO2, before any reaction takes place)
3) The NO2¯ concentration is 2/3 of its original concentration (going from 40.0 cm3 to 60.0 cm3).
2/3 x 0.5 = 0.333 M NO2¯ (the initial concentration of the NO2¯, before any reaction takes place)
4) HNO2 ionizing will result in more nitrite ion and less nitrous acid when compared to the initial concentrations. The new concentrations will be these:
[HNO2] = 0.333 − x[NO2¯] = 0.333 + x
[H3O+] = x
5) However, we make the assumption that x is small compared to 0.333 and we drop the 'subtract x' and the 'add x' from our values. We are now ready to place values into the Ka expression:
[H3O+][NO2¯] Ka = ––––––––––– [NO2]
(x) (0.333) 4.57 x 10¯4 = ––––––––– 0.333
6) Since the HNO2 and NO2¯ concentrations are the same, they will cancel out leaving the hydrogen ion concentration equal to the Ka:
x = 4.57 x 10¯4 M <--- this is the [H3O+]
7) Calculate pH:
pH = −log [H+]pH = −log 4.57 x 10¯4
pH = 3.34
Note that the pH of the solution is equal to the pKa.
Blanket Statement
Almost without exception, problems that use the Henderson Hasselbalch Equation will be solved using initial concentrations (or mole amounts). The decrease in a concentration (the 'subtract x') or the increase in a concentration (the 'add x') – relative to the initial concentrations – is usually so small that they can be ignored.
Remember, Ka value are small, which indicates that a weak acid or base only ionizes a small amount in solution. The fact that Ka values are small tells us that the reaction proceeds only a small amount to the right, with only small changes in the initial concentrations.
This "approximate" technique is so ubiquitous that many instructions and textbooks do not even mention that Henderson-Hasselbalch answers are approximations. They just solve the problems.
And, always keep in mind that, due to the experimental error (up to 5%) in the determination Ka values, there is no way to arrive at a "one true answer" status.
Below, I will sometimes mention the approximate idea, but usually I will ignore it and just do the calculation.
Problem #2: Calculate the pH of a solution prepared by mixing 15.0 mL of 0.100 M NaOH and 30.0 mL of 0.100 M benzoic acid soluion. (Benzoic acid is monoprotic; its dissociation constant is 6.46 x 10¯5.)
Solution:
Moles of NaOH = (0.100 mol/L) (0.0150 L) = 0.001500 mol
Moles of benzoic acid = (0.100 mol/L) (0.0300 L) = 0.00300 molNaOH and benzoic acid react in a 1:1 molar ratio. After reaction, we have:
moles of sodium benzoate = 0.00150 mol
moles of benzoic acid = 0.00150 molpH = pKa + log [base / acid]
pH = 4.190 + log [0.00150 / 0.00150]
pH = 4.190
Problem #3: Calculate the pH of a solution prepared by mixing 5.00 mL of 0.500 M NaOH and 20.0 mL of 0.500 M benzoic acid solution. (Benzoic acid is monoprotic: its ionization constant is 6.46 x 10¯5.)
Moles of NaOH = (0.500 mol/L) (0.00500 L) = 0.00250 mol
Moles of benzoic acid = (0.500 mol/L) (0.0200 L) = 0.00750 molNaOH and benzoic acid react in a 1:1 molar ratio. After reaction, we have:
moles of sodium benzoate = 0.00250 mol
moles of benzoic acid = 0.00750 molpH = pKa + log [base / acid]
pH = 4.190 + log [0.00250 / 0.00750]
pH = 4.190 + (−0.477)
pH = 3.713
Problem #4: How many grams of NH4Cl need to be added to 1.50 L of 0.400 M ammonia in order to make a buffer solution with pH of 8.58? Kb for ammonia is 1.77 x 10¯5
Solution:
1) Use the Henderson-Hasselbalch Equation to solve this problem:
pH = pKa + log [base / acid]
2) I already know the pH, so:
8.58 = pKa + log [base / acid]
3) I need a pKa, not a Kb or a pKb. What I need is the pKa of the ammonium ion. I get it from Kw and the Kb of ammonia:
Kw = KaKb1.00 x 10¯14 = (Ka) (1.77 x 10¯5)
Ka = 5.6497 x 10¯10
pKa = −log 5.6497 x 10¯10 = 9.24797 (I'll carry some extra digits.)
4) Add that value in:
8.58 = 9.24797 + log [base / acid]
5) Next are the moles of NH3:
(0.400 mol/L) (1.50 L) = 0.600 mol
6) Add it in:
8.58 = 9.24797 + log [0.6 / x]
7) The acid (which is the NH4Cl, specifically the NH4+ part) is our unknown.
8.58 = 9.24797 + log [0.6 / x]log [0.6 / x] = −0.66797
0.6 / x = 0.2147979
x = 2.7933234 mol
2.7933234 mol times 53.4916 g/mol = 149.42 g
to three sig figs, 149 g (that's the answer to the question)
8) We can check this by going back to the H-H Eq (and I will use molarities):
pH = 9.24797 + log [base / acid]pH = 9.24797 + log [0.400 / 1.8622156]
pH = 9.24797 + log 0.2147979
pH = 9.24797 + (−0.66797)
pH = 8.58
9) The molarity of the NH4Cl (used in step 8) came from this calc:
2.7933234 mol / 1.50 L = 1.8622156 M
Problem #5: How many grams of dry NH4Cl need to be added to 2.40 L of a 0.800 M solution of ammonia to prepare a buffer solution that has a pH of 8.90? Kb for ammonia is 1.77 x 10¯5.
Solution (uses Law of Mass Action):
1) Write the relevant chemical equation for the Ka expression:
NH4+ + H2O ⇌ H3O+ + NH3
[H3O+][NH3¯] Ka = ––––––––––– [NH4+] <--- this is the value we need to calculate
2) We need the Ka for the ammonium ion:
KaKb = Kw(Ka) (1.77 x 10¯5) = 1.00 x 10¯14
Ka = 5.6497 x 10¯10
3) We need the [H3O+]:
pH = 8.90[H3O+] = 10¯8.90
[H3O+] = 1.2589 x 10¯9 M
4) Unleash the Law of Mass Action!!!
(1.2589 x 10¯9) (0.800) 5.6497 x 10¯10 = –––––––––––––––––– x
(1.2589 x 10¯9) (0.800) x = –––––––––––––––––– 5.6497 x 10¯10 x = 1.7826079 M (that's the required ammonium concentration)
5) The mass required for 2.40 L:
(1.7826079 mol/L) (2.40 L) = mass / 53.4916 g/molmass = 228.85 g
to three sig figs, 229 g
Solution (uses Henderson-Hasselbalch Equation):
pH = pKa + log [base / acid]8.90 = 9.248 + log [0.800 / x]
log [0.800 / x] = −0.348
[0.800 / x] = 0.4487454
x = 1.782748 M (this is the required molarity of the NH4Cl
(1.782748 mol/L) (2.40 L) = mass / 53.4916 g/mol
mass = 228.8689 g
To three sig figs, 229 g
Problem #6: Determine the pH of the buffer made by mixing 0.030 mol HCl with 0.050 mol CH3COONa in 2.00 L of solution. The Ka of acetic acid is 1.77 x 10¯5.
Solution:
1) The hydrogen ion from the HCl will protonate the acetate ion according to this reaction:
H+ + CH3COO¯ ---> CH3COOHThe key point is that the reaction has a 1:1 molar ratio between the two reactants.
2) The HCl is the limiting reagent. Some CH3COOH will be made and some CH3COO¯ will be left over.
CH3COOH ---> 0.030 mol is made (as a result of H+ + CH3COO¯)
CH3COO¯ ---> 0.050 − 0.030 = 0.020 mol remains
3) The Henderson-Hasselbalch Equation is used to determine the pH:
pH = pKa + log [base / acid]pH = 4.752 + log [0.020 / 0.030]
pH = 4.752 + (−0.176) = 4.58 (to two sig figs)
Problem #7: Determine the pH of the solution after the addition of 133 mL of 0.027 M nitric acid (HNO3) to 172 mL of 0.057 M sodium hydrogen citrate (Na2C3H5O(COOH)(COO)2)
Solution:
N.B. The presence of the sodium ion will be ignored since it is a spectator ion.
1) The nitric acid is a strong acid and it will protonate the weaker acid, symbolized by HCit2¯. This is the reaction:
HCit2¯ + H+- ---> H2Cit¯ <--- this reaction goes to completion, not an equilibrium
2) If you flip the above reaction, you have the equation for the Ka2 of citric acid:
H2Cit¯ ⇌ HCit2¯ + H+ <--- this reaction is an equilibrium
3) We need to know how much HCit2¯ reacted and how much H2Cit¯ was made:
moles HNO3 ---> (0.027 mol/L) (0.133 L) = 0.003591 mol
moles HCit2¯ ---> (0.057 mol/L) (0.172 L) = 0.009804How much H2Cit¯ was made? Answer: 0.003591 mol <--- all the H+- from the nitric acid was used up
How much HCit2¯ remains? Answer: 0.009804 − 0.003591 = 0.006213 molThe Ka2 for citric acid is 1.73 x 10¯5. I found it here.
4) We now use the Henderson-Hasselbalch equation since we have a buffer:
pH = pKa + log [base / acid]pH = pKa + log (HCit2¯ / H2Cit¯)
Note: the HCit2¯ is the base since it lacks the proton. The H2Cit¯ is the acid since it has the proton that gets donated to a water molecule to make H3O+ (signified in the equation in step 2 by H+).
pH = 4.762 + log (0.006213 / 0.003591)
pH = 4.762 + 0.238
pH = 5.00 (to two sig figs)
Problem #8: 1.00 L of a buffer with a pH of 4.30 contains 0.33 M of sodium benzoate and 0.26 M of benzoic acid. What is the pH in the buffer solution after the addition of 0.058 mol of HCl to a final volume of 1.6 L?
Solution:
1) Determine pKa:
4.30 = pKa + log (0.33/0.26)4.30 = pKa + 0.104
pKa = 4.196
You can also look up pKa (or Ka) online. I decided to calculate it.
2) Determine moles after addition of HCl:
0.33 − 0.058 = 0.272 mole sodium benzoate0.26 + 0.058 = 0.318 mole benzoic acid
3) Determine new pH:
pH = 4.196 + log (0.272 / 0.318)pH = 4.196 + (−0.068)
pH = 4.13 (to two sig figs
4) Notice that the final volume of 1.60 L is not used. This is because the ratio of moles is the same as the ratio of molarities. Here is the Henderson-Hasselbalch Equation with the molarities:
pH = 4.196 + log (0.17 / 0.19875)This is true: (0.272 / 0.318) = (0.17 / 0.19875) = 0.8553459
The same answer results from the ratio of moles and the ratio of molarities.
Problem #9: A buffer solution contains 5.00 mL of 2.00 M acetic acid, 45.0 mL water and 2.05 g sodium acetate. Predict the pH of the buffer solution.
Solution:
The final volume of buffer is 50 mL (5 mL acid + 45 mL water). I will ignore significant figures until the end.
1) Calculate the molarity of just the acetic acid after it has been diluted:
M1V1 = M2V2(M1) (50 mL) = (2.0 mol/L) (5 mL)
M1 = 0.2 M
2) Calculate the molarity of the CH3COONa (molar mass = 82.03 g/mol):
moles in 2.05 g = 2.05/82.03 = 0.0250 mole in 0.050 L solutionmolarity = 0.0250 mol / 0.05 L = 0.500 M
3) Now apply the Henderson-Hasselbalch equation:
pH = pKa + log ([base] / [acid])The pKa of CH3COOH (not sodium acetate) is 4.752
pH = 4.752 + log (0.500/0.20)
pH = 4.752 + log 2.50
pH = 4.752 + 0.398
pH = 5.150 (to three sig figs)
4) You can also solve this problem by determining the number of moles of acetic acid and the number of moles of sodium acetate:
acetic acid ---> (2.0 mol/L) (0.005 L) = 0.01 molpH = 4.752 + log (0.0250 / 0.01)
pH = 4.752 + 0.398
pH = 5.150
Problem #10: We have available a 0.100 M acetic acid solution as well as a 0.100 M sodium acetic solution. What volume of each would be required to produce 1.00 L of a buffer with pH 4.000? Assume that the volumes of the two solutions are additive. (pKa = 4.752)
Solution:
pH = pKa + log [base / acid]4.000 = 4.752 + log [Ac¯ / HAc]
log [Ac¯/HAc] = −0.752
log [HAc/Ac¯] = 0.752
N.B. an acidic pH means we'll need more acid than base, so flip the ratio to put the larger value on top.
[HAc] / [Ac¯] = 5.64937 <--- carry a couple guard digits
This means you need 5.64937 times as much HAc as Ac¯
To make 1000 mL:
5.64937x + x = 1000
6.64937x = 1000
x = 150.39 mL of acetate solution used
5.64937x = 849.61 mL acetic acid solution used
N.B. I have the volumes with five sig figs but, if I made this solution, I'd label it as "pH = 4.000." In other words, three sig figs.
Bonus Problem: A solution containing 0.0158 M a diprotic acid with the formula H2A and 0.0226 M of its salt Na2A. The Ka values for the acid are 1.20 x 10¯2 (Ka1) and 5.37 x 10¯7 (Ka2). What is the pH of the solution?
Solution:
1) Write the two Ka equations:
H2A ⇌ HA¯ + H+ Ka1 HA¯ ⇌ A2¯ + H+ Ka2
2) Reverse the second equilibrium:
H2A ⇌ HA¯ + H+ Ka1 A2¯ + H+ ⇌ HA¯ 1/Ka2
3) Add the two chemical equations:
H2A + A2¯ ⇌ 2HA¯ K' = Ka1 / Ka2
4) Solve for K':
K' = 1.20 x 10¯2 / 5.37 x 10¯7 = 22346What that means is that the equilibrium VERY much favors the products (in this problem, that is the HA¯).
5) What we now do is treat this as a limiting reagent problem:
H2A + A2¯ ⇌ 2HA¯Assume 1.00 L of solution is present.
moles H2A ---> 0.0158
moles A2¯ ---> 0.0226The 1:1 stoichiometry of H2A reacting with A2¯ means H2A (the lower amount at 0.0158) is the limiting reagent.
6) Determine amount of HA¯ and H2A in solution:
A2¯ ---> 0.0226 − 0.0158 = 0.0068
HA¯ ---> (2) (0.0158) = 0.0316Reaction stoichiometry: two HA¯ produced for every one H2A used.
7) Calculate the pH as a buffer centered around Ka2
pH = 6.270 + log (0.0068 / 0.0316) = 5.603
| Fifteen Buffer Examples | Buffer Problems 11-20 | Buffer Problems 21-30 | Buffer Problems 31-40 |
| Intro. to the Henderson-Hasselbalch Equation | Return to the Acid Base menu | ||