Problems #1 - 15

All examples and problems, no solutions

Return to twelve examples of weak acid/base titrations

The first six problems are multi-part. Problems 7-15 are single calculations.

In all cases below, the percent ionization is small enough that initial concentrations can be used in place of equilibrium concentrations. This is known as the "5% rule."

**Problem #1:** Acetic acid is a weak monoprotic acid with K_{a} = 1.77 x 10¯^{5}. NaOH(s) was gradually added to 1.00 L of 0.0179 M acetic acid. (Assume no volume change occurs.)

(a) Calculate the pH of the solution after the addition of 0.0107 mol of NaOH(s).

(b) Calculate the pH of the solution after the addition of 0.0179 mol of NaOH(s).

(c) Calculate the pH of the solution after the addition of 0.0286 mol of NaOH(s).

**Solution to (a):**

1) Determine moles of acetic acid:

(0.0179 mol/L) (1.00 L) = 0.0179 mol

2) Determine moles of acetic acid and sodium acetate:

acetic acid ---> 0.0179 mol − 0.0107 mol = 0.0072 mol

sodium acetate ---> 0.0107 molThe sodium acetate results from the 1:1 molar ratio between acetic acid and NaOH when they react. 0.0107 mol of acetic acid was used up and 0.0107 mol of sodium acetate resulted.

3) Use the Henderson-Hasselbalch Equation:

pH = pK_{a}+ log [base / acid]pH = 4.752 + log (0.0107 / 0.0072)

pH = 4.752 + 0.172 = 4.924

Remember, sodium acetate is the salt of a weak acid. It is the base in the H-H Equation.

**Solution to (b):**

The reaction between acetic acid and NaOH is a 1:1 molar ratio. 0.0179 mol of acetic acid and 0.0179 mol of NaOH react, producing 0.0179 mol of sodium acetate. Because the reaction takes place in 1.00 L, the acetate concentration is 0.0179 mol/L.

This problem requires to determine the pH of a solution of the salt of a weak acid, when given the K_{a} of the weak acid. The Henderson-Hasselbalch equation is not used.

1) The K_{b} of sodium acetate is required:

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

_{b}= 1.00 x 10¯^{14}/ 1.77 x 10¯^{5}K

_{b}= 5.65 x 10¯^{10}

2) The chemical reaction for acetate reacting as a base and the K_{b} expression:

Ac¯ + H_{2}O ⇌ HAc + OH¯K

_{b}= ([HAc] [OH¯]) / [Ac¯]

3) Solve the K_{b} expression for the hydroxide ion concentration:

5.65 x 10¯^{10}= [(x) (x)] / 0.0179x = 0.00000318 M

4) Determine the pOH, then the pH:

pOH = −log 0.00000318 = 5.498pH = 14.000 − 5.498 = 8.502

**Solution to (c):**

The NaOH added is in excess. All the acetic acid is converted to sodium acetate (a weak base) and there is left-over NaOH (a strong base). In a situation like this, the presence of the strong base overwhelms any contribution by the weak base (the acetate anion).

The consequence is that the weak base is ignored and the pH calculation is based only on the strong base.

1) Calculate the hydroxide concentration remaining:

0.0286 mol − 0.0179 mol = 0.0107 mol0.0107 mol / 1.00 L = 0.0107 M

2) Calculate the pOH, then the pH:

pOH = −log 0.0107 = 1.9706pH = 14 − 1.9706 = 12.029 (to three sig figs)

**Problem #2:** Consider the titration of 30.0 mL of 0.166 M of KX with 0.154 M HCl. The pK_{a} of HX = 8.090. Give all pH values to within 0.001.

(a) What is the pH of the original solution before addition of any acid?

(b) How many mL of acid are required to reach the equivalence point?

(c) What is the pH at the equivalence point?

(d) What is the pH of the solution after the addition of 22.3 mL of acid?

(e) What is the pH of the solution after the addition of 38.8 mL of acid?

**Solution to (a):**

1) The relevant chemical equation is this:

X¯ + H_{2}O ⇌ HX + OH¯We know that KX is the salt of a weak acid (whose formula is HX). We know HX is a weak acid because it has a K

_{a}value. KX, being the salt of a weak acid, will form a basic solution.

2) We need to determine the K_{b} of X¯:

pK_{a}of HX = 8.090therefore, pK

_{b}of X¯ = 14 − 8.090 = 5.910K

_{b}= 10¯^{5.910}= 1.23 x 10¯^{6}

3) Use the K_{b} to determine the hydroxide concentration:

[HX] [OH¯] K _{b}=–––––––––– [X¯]

(x) (x) 1.23 x 10¯ ^{6}=–––––– 0.166 x = 0.0004519 M

4) Determine the pOH, then the pH:

pOH = −log 0.0004519 = 3.345pH = 14 − 3.345 = 10.655

**Solution to (b):**

1) We need to know how many moles of KX are present:

(0.166 mol/L) (0.0300 L) = 0.00498 mol

2) HCl and KX react in a 1:1 molar ratio. Therefore:

0.00498 mol of HCl is required

3) The volume of HCl is:

0.00498 mol / 0.154 mol/L = 0.0323 L = 32.3 mLThe 32.3 value plays a role in a slight difference in answers in the two ways to solve part (e) below.

**Solution to (c):**

1) This chemical reaction neutralizes all the X¯:

HCl + KX ---> HX + KCl <--- complete molecular equationH

^{+}+ X¯ ---> HX <--- net ionic, note that all the H^{+}came from the HCl

2) Then, the HX dissociates a little bit (remember, it's a weak acid):

HX ---> H^{+}+ X¯ <--- the H^{+}comes from the HX dissociatingWe will use the K

_{a}expression to solve for the H+ and then on to the pH.

3) This is the molarity of the HX solution:

0.00498 mol / 0.06234 L = 0.0798845 MThe 0.06234 L came from 30.0 mL + 32.34 mL (using one guard digit on the 32.34)

4) Determine the value for the K_{a}:

K_{a}= 10¯^{8.090}= 8.1283 x 10¯^{9}

5) Use the K_{a} expression to solve for the [H^{+}]:

[H ^{+}] [X¯]K _{a}=–––––––– [HX]

(x) (x) 8.1283 x 10¯ ^{9}=––––––––– 0.0798845 x = 0.000025482 M

6) The pH

pH = −log 0.000025482 = 4.594

**Solution to (d):**

We know that the volume of acid added is not sufficient to get to the equivalence point. That means that the HCl will run out and the solution produced will have appreciable amounts of HX and X¯. Since this is a solution of a weak acid and its salt, we have a buffer. The Henderson-Hasselbalch Equation will be used to solve this problem.

1) Determine the moles of HCl added:

(0.154 mol/L) (0.0223 L) = 0.0034342 mol

2) Determine moles of HX formed and of X¯ remaining:

X¯ remaining ---> 0.00498 mol − 0.0034342 mol = 0.0015458 mol

HX formed ---> 0.0034342 mol

3) Use the Henderson-Hasselbalch Equation to determine the pH:

pH = 8.090 + log (0.0015458 / 0.0034342)pH = 8.090 + (−0.347)

pH = 7.743

**Solution to (e):**

1) Since we calculated the mL of acid required for the equivalence point (32.3 mL), we know that the 38.8 mL of HCl means there will be excess HCl. How much excess?

38.8 − 32.34 = 6.46 mL <--- see comment below

2) What this problem becomes is to determine the pH of some strong acid in solution. There is HX present, but its contribution to the pH has been suppressed by the presence of the strong acid.

3) We need the molarity of the 6.46 mL of HCl that got diluted to 68.8 mL (the total volume of the solution).

M_{1}V_{1}= M_{2}V_{2}(0.154 mol/L) (6.46 mL) = (x) (68.8 mL)

x = 0.01446 M

4) Determine the pH:

pH = −log 0.01446 = 1.840

**Alternate solution to (e):**

1) Moles of HCl added:

(0.154 mol/L) (0.0388 L) = 0.0059752 mol

2) Moles of HCl remaning after reaction with X¯

0.0059752 mol − 0.00498 mol = 0.0009952 mol

3) New molarity of HCl:

0.0009952 mol / 0.0688 L = 0.014465 M

4) pH:

pH = −log 0.014465 = 1.840Comment: I initially used 32.3 mL in the first solution to (e) but, when I did the alternate solution, I saw I had a discrepancy of about 0.003 in the pH. The cause for that was due to rounding off in the 32.3 value. So, I used one guard digit on the 32.3, resulting in the 32.34 I used in the calculation.

**Problem #3:** A 25.0-mL solution of 0.100 M acetic acid is titrated with a 0.200 M NaOH solution. Calculate the pH after the following additions of the NaOH solution:

(a) 0.0 mL

(b) 5.00 mL

(c) 10.0 mL

(d) 12.5 mL

(e) 15.0 mL

Hint: a, d, and e are not buffers. b and c are buffers.

**Solution to part (a):**

1) Insert values into the K_{a} expression for acetic acid. The K_{a} for acetic acid is 1.77 x 10¯^{5}.

(x) (x) 1.77 x 10¯ ^{5}=–––––– 0.100 x = 1.3304 x 10¯

^{3}MpH = 2.876

**Solution to part (b):**

1) Calculate moles of acid and base in solution before reaction:

CH_{3}COOH ---> (0.100 mol/L) (0.0250 L) = 0.00250 mol

NaOH ---> (0.200 mol/L) (0.00500 L) = 0.00100 mol

2) Determine amounts of acid and acetate ion after reaction:

CH_{3}COOH ---> 0.00250 mol − 0.00100 mol = 0.00150 mol

CH_{3}COONa ---> 0.00100 mol

3) Use the Henderson-Hasselbalch equation to determine the pH of the buffer solution:

The pK_{a}of acetic acid is 4.752.

0.00100 pH = 4.752 + log ––––––– 0.00150 pH = 4.752 + (−0.176)

pH = 4.576

**Solution to part (c):**

1) Calculate moles of acid and base in solution before reaction:

CH_{3}COOH ---> (0.100 mol/L) (0.0250 L) = 0.00250 mol

NaOH ---> (0.200 mol/L) (0.01000 L) = 0.00200 mol

2) Determine amounts of acid and acetate ion after reaction:

CH_{3}COOH ---> 0.00250 mol − 0.00200 mol = 0.00050 mol

CH_{3}COONa ---> 0.00200 mol

3) Use the Henderson-Hasselbalch equation to determine the pH of the buffer solution:

pK_{a}= 4.752

0.00200 pH = 4.752 + log ––––––– 0.00050 pH = 4.752 + (0.602)

pH = 5.354

**Solution to part (d):**

1) Calculate moles of acid and base in solution before reaction:

CH_{3}COOH ---> (0.100 mol/L) (0.0250 L) = 0.00250 mol

NaOH ---> (0.200 mol/L) (0.01250 L) = 0.00250 mol

2) Determine amounts of acid and acetate ion after reaction:

CH_{3}COOH ---> 0.00250 mol − 0.00250 mol = 0 mol

CH_{3}COONa ---> 0.00250 mol

Comment: this solution is the salt of a weak acid and will wind up with a basic pH. Since it is not a buffer, the H-H equation will not be used.

3) Various calculations to prepare:

0.00250 mol / 0.0375 L = 0.066667 MI need the concentration of the sodium acetate since I will not be using a ratio as in parts b and c

K

_{a}of acetic acid is 1.77 x 10 x 10¯^{5}. I need the K_{b}of the acetate ion:K

_{a}K_{b}= K_{w}(1.77 x 10 x 10¯

^{5}) (K_{b}) = 1.00 x 10¯^{14}K

_{b}= 5.6497 x 10¯^{10}

[HAc] [OH¯] K _{b}=––––––––––– [Ac¯]

(x) (x) 5.6497 x 10¯ ^{10}=–––––––– 0.066667 x = 0.0000061372 M

The above represents the hydrolysis of the salt of a weak acid in aqueous solution.

pOH = −log 0.0000061372 = 5.212

pH = 14 − 5.212 = 8.788

**Solution to part (e):**

1) Calculate moles of acid and base in solution before reaction:

CH_{3}COOH ---> (0.100 mol/L) (0.0250 L) = 0.00250 mol

NaOH ---> (0.200 mol/L) (0.0150 L) = 0.00300 molThere is now excess NaOH.

2) Determine amounts of acid and acetate ion after reaction:

CH_{3}COOH ---> 0.00250 mol − 0.00250 mol = 0 mol

CH_{3}COONa ---> 0.00250 molNaOH remaining ---> 0.00300 mol − 0.00250 mol = 0.000500 mol

Comment: NaOH is a strong base and sodium acetate is a weak base. The presence of the strong base overwhelms the weak base, resulting in the strong base only affecting the pH.

3) Determine the molarity of the NaOH and thence to the pH:

0.000500 mol / 0.0400 L = 0.0125 MpOH = −log 0.0125 = 1.903

pH = 14 − 1.903 = 12.097

**Problem #4:** Calculate the pH of the solution in each step list below for the titration of 500. mL of 0.0100 M acetic acid (pK_{a} = 4.752) with 0.0100 M KOH

(a) after 0 mL of the titrant have been added.

(b) after 250. mL of the titrant have been added.

(c) after 490. mL of the titrant have been added.

(d) after 500. mL of the titrant have been added.

(e) after 510. mL of the titrant have been added.

(f) after 750. mL of the titrant have been added.

**Solution to (a):**

This part of the question is simply asking the pH of a solution of a weak acid. We will calculate the H^{+} of the solution and get the pH from that value.

(x) (x) 1.77 x 10¯ ^{5}=–––––––– 0.0100 x = 0.0004207137 M

pH = −log 0.0004207137 = 3.376

**Solution to (b):**

Here we have created a buffer because some of the acetic acid has been neutralized. We will use the Henderson-Hasslebalch Equation to solve this problem. Before that, however, we need to determine the moles of acetic acid remaining and the moles of potassium acetate formed.

1) Moles present before reaction:

acetic acid ---> (0.0100 mol/L) (0.500 L) = 0.00500 mol

KOH ---> (0.0100 mol/L) (0.250 L) = 0.00250 mol

2) Moles present after reaction:

acetic acid ---> 0.00500 mol − 0.00250 = 0.00250 mol

potassium acetate ---> 0.00250 mol

3) Now, the H-H Equation comes into play:

[base] pH = pK _{a}+ log––––– [acid]

0.00250 pH = 4.752 + log ––––––– 0.00250 pH = 4.752 + log 1

pH = 4.752

This is the half-equivalence point. The equivalence point will come in part (d).

**Solution to (c):**

We still have a buffer, because not all of the acid has been neutralized. So, Henderson-Hasselbalch to the rescue!

1) Moles present before reaction:

acetic acid ---> (0.0100 mol/L) (0.500 L) = 0.00500 mol

KOH ---> (0.0100 mol/L) (0.490 L) = 0.00490 mol

2) Moles present after reaction:

acetic acid ---> 0.00500 mol − 0.00490 = 0.00010 mol

potassium acetate ---> 0.00490 mol

3) Now, the H-H Equation comes into play:

0.00490 pH = 4.752 + log ––––––– 0.00010 pH = 4.752 + log 49

pH = 6.442

**Solution to (d):**

1) Moles present before reaction:

acetic acid ---> (0.0100 mol/L) (0.500 L) = 0.00500 mol

KOH ---> (0.0100 mol/L) (0.500 L) = 0.00500 mol

2) Moles present after reaction:

acetic acid ---> 0.00500 mol − 0.00500 = 0 mol

potassium acetate ---> 0.00500 mol

3) We have a solution of a salt of a weak acid. This is not a buffer.

Ac¯ + H_{2}O ⇌ HAc + OH¯

[HAc] [OH¯] K _{b}=––––––––––– [Ac¯]

4) We need the K_{b} for the acetate anion:

K_{a}K_{b}= K_{w}(1.77 x 10¯

^{5}) (K_{b}) = 1.00 x 10¯^{14}K

_{b}= 5.64972 x 10¯^{10}

5) Solve for the hydroxide ion concentration:

(x) (x) 5.64972 x 10¯ ^{10}=–––––––– 0.00500 x = 1.680732 x 10¯

^{6}MBy the way, the concentration of the potassium acetate comes from 0.00500 mol / 1.00 L.

6) Get the pOH, then the pH:

pOH = −log 1.680732 x 10¯^{6}= 5.7745pH = 14 − 5.7745 = 8.225 (to three significant figures)

**Solution to (e):**

There is now more KOH present than acetic acid. The acid will react 100% and the left-over KOH will suppress the hydrolysis of the acetate ion Only the KOH will determine the pH.

1) Moles present before reaction:

acetic acid ---> (0.0100 mol/L) (0.500 L) = 0.00500 mol

KOH ---> (0.0100 mol/L) (0.510 L) = 0.00510 mol

2) Moles present after reaction:

acetic acid ---> 0.00500 mol − 0.00500 = 0 mol3) Calculate the pOH, then the pH:

potassium acetate ---> 0.00500 mol

KOH ---> 0.00510 − 0.00500 = 0.000100 mol

molarity of KOH ---> 0.000100 mol / 1.01 L = 0.0000990099 MpOH = −log 0.0000990099 = 4.004321

pH = 14 − 4.004321 = 9.996

**Solution to (f):**

molarity of KOH ---> 0.00250 mol / 1.25 L = 0.00200 MpOH = −log 0.00200 = 2.69897

pH = 11.301

**Problem #5:** A 108.6 mL sample of 0.100 M methylamine (CH_{3}NH_{2}, K_{b} = 4.4 x 10¯^{4}) is titrated with 0.250 M HNO_{3}. Calculate the pH after the addition of each of the following volumes of acid.

(a) 0.00 mL

(b) 21.72 mL

(c) 43.44 mL

(d) 65.20 mL

**Solution to (a):**

1) This is a calculation for the pH of a weak base and is not a buffer calculation. I will use the K_{b} expression:

[CH _{3}NH_{3}^{+}] [OH¯]K _{b}=––––––––––––––– [CH _{3}NH_{2}]

(x) (x) 4.4 x 10¯ ^{4}=–––––– 0.100 x = 0.00663325 M <--- this is the hydroxide concentration

2) pOH, then pH:

pOH = −log 0.00663325 = 2.18pH = 14 − 2.18 = 11.82

**Solution to (b):**

1) Determine moles of HNO_{3} added:

(0.250 mol/L) (0.02172 L) = 0.00543 mol

2) Determine moles of CH_{3}NH_{2} present before reaction:

(0.100 mol/L) (0.1086 L) = 0.01086 mol

3) Determine moles of methyl amine remaining and moles of methyl ammonium formed after the reaction is complete:

CH_{3}NH_{2}---> 0.01086 mol − 0.00543 mol = 0.00543 mol

CH_{3}NH_{3}^{+}= 0.00543 mol

Note: see how the moles of the two components of the buffer are equal? This shows it to be a special case in titration called the half-equivalence point. At this point, the pH equals the pK_{a}.

4) Do the Henderson-Hasselbalch:

pH = pK_{a}+ log [base / acid]pH = 10.64 + log (0.00543 / 0.00543)

pH = 10.64

Note: to get the pK_{a}, I first converted the K_{b} to the pK_{b} and then subtracted it from the pK_{w} value of 14.

**Solution to (c):**

Did you notice that the volume of acid in (c) is exactly double that in (b)? The calculation in (b) was for the half-equivalence point pH, the calculation in (c) is for the equivalence point pH.

1) Determine moles of acid delivered:

(0.250 mol/L) (0.04344 L) = 0.01086 mol

2) Determine moles of base in 108.6 mL:

(0.100 mol/L) (0.1086 L) = 0.01086 mol

3) 100% of the base is neutralized by the acid leaving 0.01086 mol of the salt in 0.15204 L of solution. Calculate the molarity of the salt solution:

0.01086 mol / 0.15204 L = 0.07143 M

4) The salt of a weak base is an acid, so we need the K_{a} of CH_{3}NH_{3}^{+}:

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

_{a}) (4.4 x 10¯^{4}) = 2.2727 x 10¯^{11}

5) Here is the chemical reaction of interest:

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

6) Now, a standard K_{a} calculation:

2.2727 x 10¯^{11}= [(x) (x)] / 0.07143x = 0.00000127413 M

7) pH:

pH = −log 0.00000127413 = 5.89

**Solution to (d):**

Based on above calculations, we know we have an excess amount of acid in part (d). Some of the acid will react with the methyl amine and we will wind up with a mixture of a strong acid and a weak acid (the methyl ammonium). The presence of the strong acid will suppress any H^{+} contribution from the methyl ammonium, so the calculation will be to determine the pH of a strong acid.

1) Determine moles of acid added:

(0.250 mol/L) (0.06520 L) = 0.0163 mol

2) Determine moles of acid remaining after reaction:

0.0163 mol − 0.01086 mol = 0.00544 mol

3) Determine molarity of the acid:

0.00544 mol / 0.1738 L = 0.03130 M

4) Determine pH:

pH = −log 0.03130 = 1.50

You can solve part (d) by subtracting volumes (65.20 − 43.44) to get the volume of unreacted strong acid and using M_{1}V_{1} = M_{2}V_{2} to get the new molarity. I did it in problem 2e, so take a look there for the rest of the technique.

**Problem #6:** Calculate the pH for each of the cases in the titration of 25.0 mL of 0.210 M pyridine, C_{5}H_{5}N(aq) with 0.210 M HBr(aq). The pK_{b} of pyridine is 8.77.

(a) before addition of any HBr

(b) after addition of 12.5 mL of HBrÂ

(c) after addition of 21.0 mL of HBr

(d) after addition of 25 mL of HBr

(e) after addition of 31.0 mL of HBr

**Solution to (a):**

This is a K_{b}problem. Write the chemical equation, then the K_{b}expression:C

_{5}H_{5}N + H_{2}O ⇌ C_{5}H_{5}NH^{+}+ OH¯

[C _{5}H_{5}NH^{+}] [OH¯]K _{b}=––––––––––––––– [C _{5}H_{5}N]

(x) (x) 1.698 x 10¯ ^{9}=–––––– <--- K _{b}from 10¯^{8.77}0.210 x = 0.00001888 M

pOH = −log 0.00001888 = 4.724

pH = 14 − 4.724 = 9.28 (to two sig figs)

**Solution to (b):**

This is the half-equivalence point, but let's act like we don't know that.mmol pyridine before reaction ---> (0.210 mmol/mL) (25.0 mL) = 5.25 mmol

mmol HBr before reaction ---> (0.210 mmol/mL) (12.5 mL) = 2.625 mmolThe two react in a 1:1 ratio:

mmol pyridine after reaction ---> 5.25 mmol − 2.625 mmol = 2.625 mmol

mmol pyridinium after reaction ---> 2.625 mmolWrite the Henderson-Hasselbalch Equation:

pH = pK

_{a}+ log [base / acid]pH = 5.23 + log (2.625 / 2.625)

pH = 5.23

**Solution to (c):**

mmol pyridine before reaction ---> (0.210 mmol/mL) (25.0 mL) = 5.25 mmol

mmol HBr before reaction ---> (0.210 mmol/mL) (21.0 mL) = 4.41 mmolThe two react in a 1:1 ratio:

mmol pyridine after reaction ---> 5.25 mmol − 4.41 mmol = 0.84 mmol

mmol pyridinium after reaction ---> 4.41 mmolWrite the Henderson-Hasselbalch Equation:

pH = 5.23 + log [0.84 / 4.41]

pH = 5.23 + (−0.72) = 4.51

**Solution to (d):**

This is the equivalence point.mmol pyridine before reaction ---> (0.210 mmol/mL) (25.0 mL) = 5.25 mmol

mmol HBr before reaction ---> (0.210 mmol/mL) (25.0 mL) = 5.25 mmolThe two react in a 1:1 ratio:

mmol pyridine after reaction ---> 5.25 mmol − 5.25 mmol = 0 mmol

mmol pyridinium after reaction ---> 5.25 mmolThe chemical reaction:

C

_{5}H_{5}NH^{+}+ H_{2}O ⇌ C_{5}H_{5}N + H_{3}O^{+}We will do a K

_{a}calculation. For that, we need the molarity of the pyridinium ion:5.25 mmol / 50.0 mL = 0.105 M

[C _{5}H_{5}N] [H_{3}O^{+}]K _{a}=––––––––––––––– [C _{5}H_{5}N]

(x) (x) 5.888 x 10¯ ^{6}=–––––– 0.105 x = 0.0007863 M

pH = −log 0.0007863 = 3.10

**Solution to (e):**

The HBr will be in excess and will be the only contributor to the pH. The presence of the pyridinium ion may be ignored.The fact that the molarities of the two reactants are equal means that 25.0 mL of the pyridine was neutralized, leaving 6.00 mL of the HBr unreacted.

Use M

_{1}V_{1}= M_{2}V_{2}(0.210 mmol/mL) (6.00 mL) = (x) (56.0 mL)

x = 0.0225 M

pH = −log 0.0225 = 1.65

**Problem #7:** What is the pH at the equivalence point in the titration of 0.250 M HX (K_{a} = 5.2 x 10¯^{6}) with 0.250 M KOH?

**Discussion:**
Notice that no volumes are given, but this is not a problem. The molarities of the reactants, HX and KOH, both equal 0.250 M. At the equivalence point, the reactant moles must be equal, which means that the volumes of the two reactant solutions will need to be equal. Therefore, the total volume available to the initial HX doubles, so its concentration will be cut in half, and so too concentration of the product after neutralization by KOH. In this case, the product is KX with a concentration of 0.125 M (which is one-half of the 0.250 M of each reactant).

The end result of this is that we can assume ANY starting amount we want. The concentration of the resulting KX solution will always be 0.125 M.

**Solution:**

1) Determine concentration of KX at the equivalence point:

Assume 1.00 L of 0.250 M HX reacts with 1.00 L of 0.250 M KOH.This produces 2.00 L of solution containing 0.250 mol of KX.

[KX] = 0.250 mol / 2.00 L = 0.125 M

2) Since KX is the salt of a weak acid, we need the K_{b} of X¯

K_{a}K_{b}= K_{w}(5.2 x 10¯

^{6}) (x) = 1.0 x 10¯^{14}x = 1.923 x 10¯

^{9}

3) Solve for [OH¯] in the following reaction:

X¯ + H_{2}O ⇌ HX + OH¯

(x) (x) 1.923 x 10¯ ^{9}=–––––– 0.125 x = 1.55 x 10¯

^{5}M

4) Determine the pH:

pOH = −log [OH¯] = −log 1.55 x 10¯^{5}= 4.81pH = 14.00 − 4.81 = 9.19

**Problem #8:** What is the pH at the equivalence point in the titration of 100.0 mL of 0.100 M HCN (K_{a} = 4.9 x 10¯^{10}) with 0.100 M NaOH?

**Solution:**

1) Calculate the [NaCN] at the equivalence point:

The acid and base react in a 1:1 ratio. Since the initial concentration of the acid and base are the same, equal volumes of the acid and base are required.This means, since the final volume doubles, that the [NaCN] is half that of the concentrations of the acid and the base.

[NaCN] = 0.050 M

2) Since NaCN is the salt of a weak acid, we need the K_{b} of CN¯

K_{a}K_{b}= K_{w}(4.9 x 10¯

^{10}) (x) = 1.0 x 10¯^{14}x = 2.041 x 10¯

^{5}

3) Solve for [OH¯] in the following reaction:

CN¯ + H_{2}O ⇌ HCN + OH¯

(x) (x) 2.041 x 10¯ ^{5}=–––––– 0.050 x = 1.01 x 10¯

^{3}M

4) Determine the pH:

pOH = −log [OH¯] = −log 1.01 x 10¯^{3}= 3.00pH = 14.00 − 3.00 = 11.00

**Problem #9:** What is the pH when 25.00 mL of 0.20 M CH_{3}COOH (K_{a} = 1.77 x 10¯^{5}) has been titrated with 40.0 mL of 0.10 M NaOH?

**Solution:**

1) Determine moles of acid and base before reaction:

CH_{3}COOH: (0.20 mol/L) (0.02500 L) = 0.0050 mol

NaOH: (0.10 mol/L) (0.04000 L) = 0.0040 mol

2) Determine moles of acid and salt after reaction:

CH_{3}COOH: 0.0050 mol − 0.0040 mol = 0.0010 mol

CH_{3}COONa: 0.0040 mol

3) Determine pK_{a} for acetic acid using the K_{a} provided in the problem statement. Then, use the Henderson-Hasselbalch Equation to determine the pH.

0.0040 pH = 4.572 + log –––––– 0.0010 pH = 4.572 + 0.602

pH = 5.354

**Problem #10:** Calculate the pH at the equivalence point for the titration of 0.220 M methyl amine (CH_{3}NH_{2}) with 0.220 M HCl. The K_{b} of methylamine is 4.4 x 10¯^{4}.

**Solution:**

1) The molarity of the CH_{3}NH_{3}^{+} solution at the equivalence point will be 0.110 M. How do I know this?

CH_{3}NH_{2}+ H^{+}---> CH_{3}NH_{3}^{+}There is a 1:1 stoichiometric ratio between the methyl amine and the HCl reacting.

Since the molarities are equal, this means that equal volumes of the two solutions must be mixed in order to obtain neutralization of the CH

_{3}NH_{2}with no H^{+}left over.The CH

_{3}NH_{3}^{+}is in a solution with twice the volume, since one volume each of base and acid were mixed.This cuts the molarity in half.

2) We need the K_{a} of CH_{3}NH_{3}^{+}:

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

_{a}) (4.4 x 10¯^{4}) = 1.0 x 10¯^{14}K

_{a}= 2.2727 x 10¯^{11}

3) Now, a standard K_{a} calculation:

2.2727 x 10¯^{11}= [(x) (x)] / 0.110x = 0.00000158114 M

4) pH:

pH = −log 0.00000158114 = 5.80

**Problem #11:** 25.0 mL of 0.10 M acetic acid (HAc) is titrated with 0.10 M NaOH. What is the pH at the equivalence point?

**Solution:**

1) We know the following:

HAc + OH¯ ---> Ac¯ + H_{2}O25.0 mL of NaOH is required to reach equivalence.

The total volume of the solution is 50.0 mL.

The moles of sodium acetate are 0.0025 mol

2) Calculate the molarity of the sodium acetate:

0.0025 mol / 0.050 L = 0.0050 M

3) Calculate the K_{b} of sodium acetate:

K_{a}K_{b}= K_{w}(1.77 x 10¯

^{5}) (K_{b}) = 1.00 x 10¯^{14}K

_{b}= 5.65 x 10¯^{10}

4) Calculate pH of the solution:

(x) (x) 5.65 x 10¯ ^{10}=–––––– 0.0500 x = 5.315 x 10¯

^{6}M (this is the hydroxide ion concentration)pOH = 5.274

pH = 8.726

**Problem #12:** A 24.60 mL sample of a 0.447 M aqueous hydrofluoric acid solution is titrated with a 0.328 M aqueous barium hydroxide solution. What is the pH at the equivalence point of this titration? The K_{a} of HF = 7.2 x 10¯^{4}.

**Solution:**

1) Determine the volume of barium hydroxide solution added:

2HF(aq) + Ba(OH)_{2}(aq) ---> BaF_{2}(aq) + 2H_{2}O(ℓ)

M _{a}V_{a}M _{b}V_{b}––––– = ––––– n _{a}n _{b}

(0.447 M) (24.60 mL) (0.328 M) (V _{b})–––––––––––––––– = –––––––––––– 2 1 (2) (0.328 M) (V

_{b}) = (1) (0.447 M) (24.60 mL)V

_{b}= 16.7625 mL

2) Determine the moles of fluoride ion in solution:

MV ---> (0.447 mol/L) (0.02460 mol) = 0.0109962 mol

3) Determine the molarity of fluoride ion in the now-titrated solution:

0.0109962 mol / 0.0413625 L = 0.26585 M

4) The reaction of interest is the hydrolysis of the fluoride ion:

F¯ + H_{2}O ⇌ HF + OH¯

5) We require the K_{b} of F¯:

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

_{b}= K_{w}/ K_{a}K

_{b}= 1.00 x 10¯^{14}/ 7.2 x 10¯^{4}K

_{b}= 1.3889 x 10¯^{11}

6) Write the K_{b} expression, substitute into it and solve:

[HF] [OH¯] K _{b}=––––––––––– [F¯]

(x) (x) 1.3889 x 10¯ ^{11}=––––––– 0.26585 x = $\sqrt{\mathrm{(1.3889\; x\; 10\xaf11)\; (0.26585)}}$

x = 0.00000192156 M (this is the hydroxide concentration)

7) Determine the pOH, then the pH:

pOH = −log 0.00000192156 = 5.72pH = 14 − 5.72 = 8.28

**Problem #13:** A 29.6 mL sample of 0.243 M dimethylamine, (CH_{3})_{2}NH, is titrated with 0.243 M hydroiodic acid. Determine the pH after adding 12.9 mL of hydroiodic acid. K_{b} for (CH_{3})_{2}NH = 5.9 x 10¯^{4}.

**Solution:**

1) Write the balanced chemical reaction:

H^{+}+ (CH_{3})_{2}NH ---> (CH_{3})_{2}NH_{2}^{+}1:1 molar ratio between the two reactants.

2) Determine moles of reactants before any reaction takes place:

HI ---> (0.243 mol/L) (0.0129 L) = 0.0031347 mol

(CH_{3})_{2}NH ---> (0.243 mol/L) (0.0296 L) = 0.0071928 mol

3) Allow the two reactants to react:

HI ---> zero amount remaining

(CH_{3})_{2}NH ---> 0.0071928 − 0.0031347 = 0.0040581 mol(CH

_{3})_{2}NH_{2}^{+}---> 0.0031347 mol produced

4) I'm going to use the Henderson-Hasselbalch Equation in the step just below, but I first need to convert the K_{b} of dimethylamine to the K_{a} of its conjugate acid (called dimethylammonium, sometimes also called dimethylazanium):

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

_{a}) (5.9 x 10¯^{4}) = 1.0 x 10¯^{14}K

_{a}= 1.6949 x 10¯^{11}

5) We use the Henderson-Hasselbalch Equation to get the pH of the solution because we now have a buffer (some weak base and some of its salt) after all of the strong acid is consumed.

[base] pH = pK _{a}+ log––––– [acid]

0.0040581 pH = 10.77 + log ––––––––– 0.0031347 pH = 10.771 + 0.112 = 10.88 (to two sig figs)

**Problem #14:** A 46.2 mL sample of a 0.392 M aqueous hydrofluoric acid solution is titrated with a 0.476 M aqueous sodium hydroxide solution. What is the pH after 22.2 mL of base have been added? The K_{a} of HF = 7.2 x 10¯^{4}.

**Solution:**

1) The chemical reaction of interest:

HF + OH¯ ---> F¯ + H_{2}O

2) The molar amounts of the two reactants before any reaction takes place:

HF ---> (0.392 mol/L) (0.0462 L) = 0.0181104 mol

NaOH ---> (0.476 mol/L) (0.0222 L) = 0.0105672 molNote: NaOH is the limiting reagent.

3) The molar amounts remaining after reaction ceases:

HF ---> 0.0181104 mol − 0.0105672 mol = 0.0075432 mol

F¯ ---> 0.0075432 mol produced

4) The Henderson-Hasselbalch Equation is used because we have a buffer:

[base] pH = pK _{a}+ log––––– [acid]

0.0075432 pH = 3.1427 + log ––––––––– 0.0075432 The division is seen to be equal to 1. The log of 1 is zero. Therefore:

pH = 3.14 (to two sig figs)

**Problem #15:** A 100. mL sample of 0.25 M CH_{3}NH_{2}(aq) is titrated with a 100. mL of 0.25 M HNO_{3}(aq). Determine the pH after the reaction is complete. (The K_{b} for methylamine is 4.4 x 10¯^{4})

**Solution:**

1) Determine millimoles of each reactant:

CH_{3}NH_{2}---> (0.25 mmol/mL) (100. mL) = 25 mmol

HNO_{3}---> (0.25 mmol/mL) (100. mL) = 25 mmol

2) Write the chemical equation for methylamine and nitric acid reacting:

CH_{3}NH_{2}+ HNO_{3}---> CH_{3}NH_{3}^{+}+ NO_{3}¯Methylamine and nitric acid react in a 1:1 molar ratio

3) Determine how much of which reactant is left over:

25.0 − 25.0 = 0Both reactants are completely consumed.

4) 25.0 mmol of methylammonium ion remains in solution. Determine its molarity:

25.0 mmol / 200. mL = 0.125 M

5) Methylammonium ion hydrolyzes. Write the chemical equation:

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

6) Write the K_{a} expression for methylammonium ion and solve for the [H_{3}O^{+}]:

[CH _{3}NH_{2}] [H_{3}O^{+}]K _{a}=––––––––––––––– [CH _{3}NH_{3}^{+}]

(x) (x) 2.2727 x 10¯ ^{11}=–––––– 0.125 x = 0.0000016855 M

7) Calculate the pH:

pH = −log 0.0000016855 = 5.77

**Bonus Problem:** A 0.987 g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 60.0 mL of this solution was titrated with 0.07600 M NaOH. The pH after the addition of 28.63 mL of base was 4.840, and the equivalence point was reached with the addition of 46.17 mL of base.

(a) What is the molar mass of the acid?

(b) What is the pK_{a}of the acid?

**Solution to (a):**

1) Moles of NaOH required to reach the equivalence point:

(0.07600 mol/L) (0.04617 L) = 0.00350892 molThis is how many moles of acid are present in the 60.0 mL sample that was titrated.

2) Moles of acid in the 100.0 mL of solution:

0.00350892 mol is to 60.0 mL as x is to 100.0 mLx = 0.0058482 mol

3) The molar mass of the acid:

0.987 g / 0.0058482 mol = 169 g/mol

**Solution to (b):**

1) Moles NaOH in 28.63 mL:

(0.07600 mol/L) (0.02863 L) = 0.00217588 mol

2) The NaOH neutralized 0.00217588 mol of the acid to form a salt (which is a weak base). How many moles of weak acid remain?

0.00350892 mol − 0.00217588 mol = 0.00133304 molThe 0.0035 value was obtained in the solution to part (a).

3) The Henderson-Hasselbalch Equation is used:

pH = pK_{a}+ log [base / acid]4.840 = pK

_{a}+ log [0.00217588 / 0.00133304]4.840 = pK

_{a}+ 0.213pK

_{a}= 4.627

All examples and problems, no solutions