Problems #1 - 10

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

**Problem #1:** 20.0 cm^{3} of 1.00 M HNO_{2} and 40.0 cm^{3} of 0.500 M NaNO_{2} are mixed. What is the pH of the resulting solution? pK_{a} 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 K_{a} expression. Just be aware that the K_{a} expression and the Henderson-Hasselbalch Equation are the same thing.

**Solution:**

1) This is the chemical reaction of interest:

HNO_{2}+ H_{2}O ⇌ H_{3}O^{+}+ NO_{2}¯

2) The HNO_{2} is diluted by the addition of NaNO_{2} solution.

M_{1}V_{1}= M_{2}V_{2}(1.00 mol/L) (20.0 cm

^{3}) = (x) (60.0 cm^{3}) = 0.333 M HNO_{2}(the initial concentration of the HNO_{2}, before any reaction takes place)

3) The NO_{2}¯ concentration is 2/3 of its original concentration (going from 40.0 cm^{3} to 60.0 cm^{3}).

2/3 x 0.5 = 0.333 M NO_{2}¯ (the initial concentration of the NO_{2}¯, before any reaction takes place)

4) HNO_{2} ionizing will result in more nitrite ion and less nitrous acid when compared to the initial concentrations. The new concentrations will be these:

[HNO_{2}] = 0.333 − x[NO

_{2}¯] = 0.333 + x[H

_{3}O^{+}] = 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 K_{a} expression:

[H _{3}O^{+}][NO_{2}¯]K _{a}=––––––––––– [NO _{2}]

(x) (0.333) 4.57 x 10¯ ^{4}=––––––––– 0.333

6) Since the HNO_{2} and NO_{2}¯ concentrations are the same, they will cancel out leaving the hydrogen ion concentration equal to the K_{a}:

x = 4.57 x 10¯^{4}M <--- this is the [H_{3}O^{+]}

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 pK

_{a}.

**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, K_{a} value are small, which indicates that a weak acid or base only ionizes a small amount in solution. The fact that K_{a} 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 K_{a} 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 = pK

_{a}+ 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 = pK

_{a}+ 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 NH_{4}Cl 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? K_{b} for ammonia is 1.77 x 10¯^{5}

**Solution:**

1) Use the Henderson-Hasselbalch Equation to solve this problem:

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

2) I already know the pH, so:

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

3) I need a pK_{a}, not a K_{b} or a pK_{b}. What I need is the pK_{a} of the ammonium ion. I get it from K_{w} and the K_{b} of ammonia:

K_{w}= K_{a}K_{b}1.00 x 10¯

^{14}= (K_{a}) (1.77 x 10¯^{5})K

_{a}= 5.6497 x 10¯^{10}pK

_{a}= −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 NH_{3}:

(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 NH_{4}Cl, specifically the NH_{4}^{+} 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 NH_{4}Cl (used in step 8) came from this calc:

2.7933234 mol / 1.50 L = 1.8622156 M

**Problem #5:** How many grams of dry NH_{4}Cl 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? K_{b} for ammonia is 1.77 x 10¯^{5}.

**Solution (uses Law of Mass Action):**

1) Write the relevant chemical equation for the K_{a} expression:

NH_{4}^{+}+ H_{2}O ⇌ H_{3}O^{+}+ NH_{3}

[H _{3}O^{+}][NH_{3}¯]K _{a}=––––––––––– [NH _{4}^{+}]<--- this is the value we need to calculate

2) We need the K_{a} for the ammonium ion:

K_{a}K_{b}= K_{w}(K

_{a}) (1.77 x 10¯^{5}) = 1.00 x 10¯^{14}K

_{a}= 5.6497 x 10¯^{10}

3) We need the [H_{3}O^{+}]:

pH = 8.90[H

_{3}O^{+}] = 10¯^{8.90}[H

_{3}O^{+}] = 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 = pK_{a}+ 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 NH

_{4}Cl(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 CH_{3}COONa in 2.00 L of solution. The K_{a} 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^{+}+ CH_{3}COO¯ ---> CH_{3}COOHThe key point is that the reaction has a 1:1 molar ratio between the two reactants.

2) The HCl is the limiting reagent. Some CH_{3}COOH will be made and some CH_{3}COO¯ will be left over.

CH_{3}COOH ---> 0.030 mol is made (as a result of H^{+}+ CH_{3}COO¯)

CH_{3}COO¯ ---> 0.050 − 0.030 = 0.020 mol remains

3) The Henderson-Hasselbalch Equation is used to determine the pH:

pH = pK_{a}+ 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 (HNO_{3}) to 172 mL of 0.057 M sodium hydrogen citrate (Na_{2}C_{3}H_{5}O(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 HCit^{2}¯. This is the reaction:

HCit^{2}¯ + H^{+-}---> H_{2}Cit¯ <--- this reaction goes to completion, not an equilibrium

2) If you flip the above reaction, you have the equation for the K_{a2} of citric acid:

H_{2}Cit¯ ⇌ HCit^{2}¯ + H^{+}<--- this reaction is an equilibrium

3) We need to know how much HCit^{2}¯ reacted and how much H_{2}Cit¯ was made:

moles HNO_{3}---> (0.027 mol/L) (0.133 L) = 0.003591 mol

moles HCit^{2}¯ ---> (0.057 mol/L) (0.172 L) = 0.009804How much H

_{2}Cit¯ was made? Answer: 0.003591 mol <--- all the H^{+-}from the nitric acid was used up

How much HCit^{2}¯ remains? Answer: 0.009804 − 0.003591 = 0.006213 molThe K

_{a2}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 = pK_{a}+ log [base / acid]pH = pK

_{a}+ log (HCit^{2}¯ / H_{2}Cit¯)Note: the HCit

^{2}¯ is the base since it lacks the proton. The H_{2}Cit¯ is the acid since it has the proton that gets donated to a water molecule to make H_{3}O^{+}(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 pK_{a}:

4.30 = pK_{a}+ log (0.33/0.26)4.30 = pK

_{a}+ 0.104pK

_{a}= 4.196You can also look up pK

_{a}(or K_{a}) 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:

M_{1}V_{1}= M_{2}V_{2}(M

_{1}) (50 mL) = (2.0 mol/L) (5 mL)M

_{1}= 0.2 M

2) Calculate the molarity of the CH_{3}COONa (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 = pK_{a}+ log ([base] / [acid])The pK

_{a}of CH_{3}COOH (not sodium acetate) is 4.752pH = 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. (pK_{a} = 4.752)

**Solution:**

pH = pK_{a}+ 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 H_{2}A and 0.0226 M of its salt Na_{2}A. The K_{a} values for the acid are 1.20 x 10¯^{2} (K_{a1}) and 5.37 x 10¯^{7} (K_{a2}). What is the pH of the solution?

**Solution:**

1) Write the two K_{a} equations:

H _{2}A ⇌ HA¯ + H^{+}K _{a1}HA¯ ⇌ A ^{2}¯ + H^{+}K _{a2}

2) Reverse the second equilibrium:

H _{2}A ⇌ HA¯ + H^{+}K _{a1}A ^{2}¯ + H^{+}⇌ HA¯1/K _{a2}

3) Add the two chemical equations:

H _{2}A + A^{2}¯ ⇌ 2HA¯K' = K _{a1}/ K_{a2}

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:

H_{2}A + A^{2}¯ ⇌ 2HA¯Assume 1.00 L of solution is present.

moles H

_{2}A ---> 0.0158

moles A^{2}¯ ---> 0.0226The 1:1 stoichiometry of H

_{2}A reacting with A^{2}¯ means H_{2}A (the lower amount at 0.0158) is the limiting reagent.

6) Determine amount of HA¯ and H_{2}A in solution:

A^{2}¯ ---> 0.0226 − 0.0158 = 0.0068

HA¯ ---> (2) (0.0158) = 0.0316Reaction stoichiometry: two HA¯ produced for every one H

_{2}A used.

7) Calculate the pH as a buffer centered around K_{a2}

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 |