Problems #1 - 10

Examples 1 to 10 | Problems 21 to 30 | |

Problems 11 to 20 | Problems 31 to 40 | Return to the Acid Base menu |

**Problem #1:** 20.0 cm^{3} of 1.00 mol HNO_{2} and 40.0 cm^{3} of 0.500 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) Common ion effect....

The presence of a common ion, in this case the nitrite ion, will reduce the ionization of the weak acid HNO_{2}(which is only observed in aqueous solution.)

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

20 mL HNO_{2}x 1.0vM / 60 mL = 0.33vM HNO_{2}

3) The NO_{2}¯ concentration is 2/3 of its original concentration.

2/3 x 0.5 = 0.33M NO_{2}¯

4) 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}, and the pH of the solution equal to the pK_{a}.

K_{a}= [H^{+}][NO_{2}¯] / [HNO_{2}]4.57 x 10¯

^{4}= [(x)(0.33M)] / 0.33Mx = 4.57 x 10¯

^{4}= [H^{+}]

5) pH = -log[H^{+}]

pH = -log(4.57x10¯^{4})pH = 3.34

**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.)Add that value in:

8.58 = 9.24797 + log [base / acid]

4) Next are the moles of NH_{3}:

(0.400 mol/L) (1.50 L) = 0.600 molAdd it in:

8.58 = 9.24797 + log [0.6 / x]

The acid (which is the NH

_{4}Cl, specifically the ammonium part) is our unknown. Note: at the end of my solution is an alternate solution path that solves for the required molarity of the NH_{4}Cl. For there you can get to the required grams very easily.

5) Solve:

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

6) 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

The molarity of the NH_{4}Cl came from this calc:

2.7933234 mol / 1.50 L

**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:**

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 0.0300 mol HCl with 0.0500 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 ---> 0.0300 mol is made (as a result of H^{+}+ CH_{3}COO¯)

CH_{3}COO¯ ---> 0.0500 - 0.0300 = 0.0200 mol remains

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

pH = pK_{a}+ log [base / acid]pH = 4.752 + log [0.0200 / 0.0300]

pH = 4.752 + 0.176 = 4.928

**Problem #7:** Determine the pH of the solution after the addition of 133 mL of 0.0270 M nitric acid (HNO_{3}) to 172 mL of 0.0570 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^{+}<--- note that 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}---> (0.0270 mol/L) (0.133 L) = 0.003591 mol

moles HCit^{2}¯ ---> (0.0570 mol/L) (0.172 L) = 0.009804How much H

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

How much HCit^{2}¯ remains? Ans: 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 = 5.000

**Problem #8:** 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 concentration of H^{+} in the 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:

0.330 - 0.058 = 0.272 mole sodium benzoate0.260 + 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 = 4.128

4) Convert pH to [H^{+}]:

[H^{+}] = 10^{-4.128}= 7.745 x 10^{-5}M

**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). We'll ignore significant figures until the end.

1) Calculate the molarity of the final acid solution:

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 = 0.0249 mole in 0.050 L solutionmolarity = 0.0249 mol / 0.05 L = 0.498 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 (0.498/0.20)

pH = 4.752 + log 2.49

pH = 4.752 + 0.396

pH = 5.148

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 (0.0249 / 0.01)

**Problem #10:** What volumes of 0.100 M acetic acid and 0.100 M sodium acetate would be required to produce 1.00 L of buffer at pH 4.000? (pK_{a} = 4.752)

**Solution:**

pH = pK_{a}+ log [Ac¯]/[HAc]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

plus 849.61 mL acetic acid

**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_{2} values for the acid are 1.20 x 10¯^{2} (K_{a2}) and 5.37 × 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 × 10¯^{7}= 22346What that means is that there is a VERY product (the HA¯) favored equilibrium.

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 stoich: 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

Examples 1 to 10 | Problems 21 to 30 | |

Problems 11 to 20 | Problems 31 to 40 | Return to the Acid Base menu |