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**Problem #1:** Consider the titration of a 24.0-mL sample of 0.105 M CH_{3}COOH with 0.130 M NaOH. What is . . .

a) the initial pH?

b) the volume of added base required to reach the equivalence point?

c) the pH at 6.00 mL of added base?

d) the pH at one-half of the equivalence point?

e) the pH at the equivalence point?

**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}.

1.77 x 10^{-5}= [(x) (x)] / 0.105x = 1.3633 x 10

^{-3}MpH = 2.865

**Solution to part b:**

1) Calculate moles of acid present:

(0.105 mol/L) (0.0240 L) = 2.52 x 10^{-3}moles

2) Determine moles of base required to react equivalence point:

CH3) Calculate volume of base solution required:_{3}COOH + NaOH ---> CH_{3}COONa + H_{2}OThere is a 1:1 molar ratio between acetic acid and sodium hydroxide.

Therefore, 2.52 x 10

^{-3}moles of base required

2.52 x 10^{-3}mol divided by 0.130 mol/L = 0.0194 L = 19.4 mL

**Solution to part c:**

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

CH_{3}COOH: 2.52 x 10^{-3}mol

NaOH: (0.00600 L) (0.130 mol/L) = 7.80 x 10^{-4}mol

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

CH_{3}COOH: 2.52 x 10^{-3}mol - 7.80 x 10^{-4}mol = 1.74 x 10^{-3}mol

CH_{3}COONa: 7.80 x 10^{-4}mol

3) Use Henderson-Hasselbalch equation to determine pH of (now) buffered solution:

pH = 4.752 + log (7.80 x 10^{-4}/ 1.74 x 10^{-3})pH = 4.752 + log 0.4483

pH = 4.404

Note that the new molarities were not calculated for the log term, rather the mole amounts were used directly. This is because both mole amounts exist in the same 30.0 mL solution. There would be identical volume amounts in the numerator and denominator of the log term, so they cancel out.

**Solution to part d:**

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount. Just below, I will use a '1' to symbolize the amount. Keep in mind that it is not the actual amount.

1) Use the Henderson-Hasselbalch Equation:

pH = 4.752 + log (1/1)pH = 4.752

You may use the actual number of moles if you wish. It simply does not matter since the log term will always zero out at the half-equivalence point.

Note: pH = pK_{a} at the half-equivalence point. Remember that. You stand a very good chance of being asked a half-equivalence point question on your test.

**Solution to part e:**

The solution is now completely composed of a salt of a weak acid. The pH of this solution will be basic.

1) Calculate molarity of sodium acetate:

2.52 x 10^{-3}mol / 0.0434 L = 0.0581 M

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

K_{w}= K_{a}K_{b}1.00 x 10

^{-14}= (1.77 x 10^{-5}) (x)x = 5.65 x 10

^{-10}

3) Calculate pH of the solution:

5.65 x 10^{-10}= [(x) (x)] / 0.0581x = 5.73 x 10

^{-6}M (this is the hydroxide ion concentration)pOH = 5.242

pH = 8.758

**Problem #2:** What is the pH when 25.00 mL of 0.20 M CH_{3}COOH 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) Use Henderson-Hasselbalch Equation to determine pH:

pH = 4.572 + log (0.0040/0.0010)pH = 4.572 + 0.602

pH = 5.354

**Problem #3:** A 0.552 g sample of ascorbic acid (Vitamin C) was dissolved in water to a total volume of 20.0 mL and titrated with 0.1103 M KOH. The equivalence point was reached at 28.42 mL and the pH of the solution at 10.0 mL of added base was 3.72. What is: (a) the molar mass of Vitamin C and (b) its K_{a}?

**Solution to part a:**

1) Determine moles of base used to reach equivalence point:

(0.1103 mol/L) (0.02842 L) = 0.003134726 mol

2) Assuming ascorbic acid is monoprotic (a necessary assumption to solve the problem!), calculate its molar mass:

0.552 g / 0.003134726 mol = 176 g/mol

**Solution to part b:**

1) Determine amounts of acid and salt at 10.0 mL of added base:

acid: 0.003134726 mol x (18.42/28.42) = 0.002031726 mol

base: 0.003134726 mol x (10.0/28.42) = 0.001103 mol

2) Use the Henderson-Hasselbalch Equation to determine the pK_{a}:

3.72 = x + log (0.001103/0.002031726)3.72 = x + (-0.265)

x = 3.98 (to two, not three, sig figs)

10

^{-pKa}yields the K_{a}

**Problem #4:** Calculate the pH at the points indicated below if 50.0 mL of 0.100 M aniline hydrochloride is titrated with 0.185 M NaOH. (K_{a} for aniline hydrochloride is 2.4 x 10^{-5})

C_{6}H_{5}NH_{3}^{+}(aq) + OH¯ (aq) ---> C_{6}H_{5}NH_{2}(aq) + H_{2}O (l)

a) before the titration begins

b) at the equivalence point

c) at the midpoint of the titration

d) after 20 mL of NaOH has been added

e) after 30 mL of NaOH has been added

**Solution to part a:**

Aniline hydrochloride (formula is C_{6}H_{5}NH_{3}^{+}Cl¯) is a salt of the weak base aniline. The salt will form an acidic solution.

C_{6}H_{5}NH_{3}^{+}(aq) ---> C_{6}H_{5}NH_{2}(aq) + H_{3}O^{+}(aq)2.4 x 10

^{-5}= [(x) (x)] / 0.100x = 0.00155 M

pH = 2.810

**Solution to part b:**

1) Calculate moles of aniline hydrochloride initially present:

(0.100 mol/L) (0.0500 L) = 0.00500 mol

2) Calculate the volume of NaOH solution required to reach equivalence point:

(0.185 mol/L) (x) = 0.00500 molx = 0.0270 L = 27.0 mL

3) Calculate new molarity of aniline:

0.00500 mol / 0.0770 L = 0.064935 M

4) Calculate K_{b} for aniline:

(2.4 x 10^{-5}) (x) = 1.00 x 10^{-14}x = 4.17 x 10

^{-10}

5) Calculate pH of solution:

4.17 x 10^{-10}= [(x) (x)] / 0.064935x = 5.20364 x 10

^{-6}MpOH = -log 5.20364 x 10

^{-6}= 5.284pH = 8.716

**Solution to part c:**

A better term for midpoint of the titration is half-equivalence point. At the half-equivalence point, exactly half the weak acid (in this case) has been titrated. The half that has been titrated has been converted into a base (in our case, named aniline). This solution is a buffer, so we use the Henderson-Hasselbalch Equation:

pH = pK_{a}+ log ([base] / [acid])However, the base/acid ratio at the half-equivalence always equals one, therefore:

pH = pK

_{a}pH = - log 2.4 x 10

^{-5}pH = 4.62

**Solution to part d:**

1) Calculate moles of each substance before reacting:

aniline hydrochloride: (0.100 mol/L) (0.0500 L) = 0.00500 mol

NaOH: (0.185 mol/L) (0.0200 L) = 0.00370 mol

2) Calculate moles of each substance after reacting:

aniline hydrochloride: 0.00500 mol - 0.00370 mol = 0.00130 mol

aniline: 0.00370 mol

3) Use the Henderson-Hasselbalch Equation:

pH = pK_{a}+ log ([base] / [acid])pH = 4.620 + log (0.00370 / 0.00130)

pH = 5.07 (to two sig figs)

**Solution to part e:**

27.0 mL of the NaOH solution goes to converting aniline hydrochloride into aniline, a weak base. 3.0 mL of 0.185 M NaOH are left over. Anytime you have a mixture of a strong base and a weak base, ignore the weak and concentrate on the strong.

1) Find new molarity of NaOH:

M_{1}V_{1}= M_{2}V_{2}(0.185 mol/L) (3.0 mL) = (x) (80.0 mL)

x = 0.0069375 M

2) Find pOH, then pH:

pOH = - log 0.0069375 = 2.159pH = 14 - pOH = 11.841

**Problem #5:** 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_{w}= K_{a}K_{b}1.00 x 10

^{-14}= (1.77 x 10^{-5}) (x)x = 5.65 x 10

^{-10}

4) Calculate pH of the solution:

5.65 x 10^{-10}= [(x) (x)] / 0.0500x = 5.315 x 10

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

pH = 8.726

**Problem #6:** 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?

**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¯K

_{b}= ([HX] [OH¯]) / [X¯]1.923 x 10

^{-9}= [(x) (x)] / 0.125x = 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 #7:** 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. Therefore, 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¯K

_{b}= ([HCN] [OH¯]) / [CN¯]2.041 x 10

^{-5}= [(x) (x)] / 0.050x = 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 #8:** If an acetate buffer solution was going to be prepared by neutralizing HC_{2}H_{3}O_{2} with 0.10 M NaOH, what volume (in mL) of 0.10 M NaOH would need to be added to 10.0 mL of 0.10 M HC_{2}H_{3}O_{2} to prepare a solution with pH = 5.50?

**Solution:**

Comment: In doing the salt (sodium acetate) and the acid (acetic acid), I'm going to use moles rather than molarity. Since everything occurs in the same volume of solution, the ratio of salt moles to acid moles is the same as the ratio of molarities. Besides, we don't know the final molarities since we are adding an unknown volume of NaOH solution.

1) We need to know the initial moles of acetic acid in the solution:

(0.010 L) (0.1 mol/L) = 0.001 mol

2) Let us insert values into the H-H equation:

pH = pKa + log (salt/acid)5.50 = 4.752 + log (x / (0.001 - x))

4.752 is the pK

_{a}of acetic acid

x is the moles of sodium acetate produced by the NaOH reacting

0.001 - x is the amount of acetic acid remaining in solution.The moles of acetate will give us moles of NaOH since there is a 1:1 molar ratio between the two.

3) Continue solving:

log (x / 0.001 - x) = 0.748

antilog both sides

(x / 0.001 - x) = 5.598

cross multiply & simplify to get:

6.598x = 5.598 x 10^{-3}x = 8.5 x 10

^{-4}moles

4) Let us determine the volume of NaOH required:

8.5 x 10^{-4}mol divided by 0.1 mol/L = 8.5 x 10^{-3}L = 8.5 mL

**Problem #9:** Someday!

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