Go to 10 weak acid/base titration problems

Examples 1 & 3 are the titration of a weak acid with a strong base.

Examples 2 & 4 are the titration of a weak base with a strong acid.

Example 5 is the titration of the salt of a weak base (which is a weak acid) with a strong base.

Example 6 is the titration of the salt of a weak acid (which is a weak base) with a strong acid.

All six examples are multi-part problems.

**Example #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}.

(x) (x) 1.77 x 10¯ ^{5}=––––––– 0.105 x = 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 / 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:

7.80 x 10¯ ^{4}pH = 4.752 + log ––––––––– 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:

1 pH = 4.752 + log ––– 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:

(x) (x) 5.65 x 10¯ ^{10}=–––––– 0.0581 x = 5.73 x 10¯

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

pH = 8.758

**Example #2:** Consider the titration of 100.0 mL of 0.200 M CH_{3}NH_{2} by 0.100 M HCl. Calculate the pH for (a) through (f). (K_{b} for CH_{3}NH_{2} = 4.4 x 10¯^{4})

(a) after 0.0 mL of HCl added.

(b) after 20.0 mL of HCl added.

(c) after 65.0 mL of HCl added.

(d) at the equivalence point.

(e) after 300.0 mL of HCl added.

(f) At what volume of HCl added does the pH = 10.643?

**Solution to (a):**

No strong acid has been added, so this solution is simply a solution of a weak base. We use the K_{b} expression to solve for the hydroxide ion concentration and thence to the pH.

1) The reaction of interest is this one:

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

2) The relevant K_{b} expression is:

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

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

(x) (x) 4.4 x 10¯ ^{4}=–––––– 0.200 x = 0.00938083 M <--- some guard digits

4) Solve for the pOH, then the pH

pOH = −log 0.00938083 = 2.02776pH = 14 − 2.02776 = 11.97

**Solution to (b):**

Some HCl has been added and some (not all) of the base has reacted. After the reaction is complete, there is a mixture of CH_{3}NH_{3}^{+} and CH_{3}NH_{2} in solution. All the HCl has been used up and we have a buffer. The Henderson-Hasselbalch Equation will be used to solve this problem.

1) The chemical reaction of interest is this:

CH_{3}NH_{2}+ H_{3}O^{+}---> CH_{3}NH_{3}^{+}+ H_{2}ONote that it is not an equilibrium, it goes to completion and all the H

^{+}(from HCl) is used up. Later on, in a later part of the problem, the HCl will be in excess.

2) How much HCl was added to the solution?

HCl ---> (0.100 mol/L) (0.0200 L) = 0.00200 mol

3) The HCl reacts with CH_{3}NH_{2} to make CH_{3}NH_{3}^{+}:

CH_{3}NH_{2}at start ---> (0.200 mol/L) (0.1000 L = 0.0200 molCH

_{3}NH_{2}after reacting with HCl ---> 0.0200 mol − 0.00200 mol = 0.018 mol

4) How much CH_{3}NH_{3}^{+} was made?

0.00200 moldue to the 1:1 stoichiometry between HCl and CH

_{3}NH_{2}

5) Enter the Henderson-Hasselbalch!

[CH _{3}NH_{2}]pH = pK _{a}+ log–––––––––– [CH _{3}NH_{3}^{+}]

0.0180 pH = 10.643 + log –––––– 0.0020 pH = 10.643 + 0.954

pH = 11.60

Notice that I did not bother to use molarities in the H-H. This is because both substances are in the same volume (120.0 mL in this case) and so their influence simply cancels out. In other words, the ratio of moles is the same exact value as the ratio of molarities.

Also take note that I used the pK_{a} of the methylammonium ion. Since it is the conjugate acid to the base (the methyl amine), the K_{a} and K_{b} are related to each other by K_{w}:

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

**Solution to (c):**

This problem is just like (b), except that 65.0 mL of HCl is added. The solution technique is the same as in (b) with just a slight change in amounts of chemical substances.

1) HCl added:

(0.100 mol/L) (0.0650 L) = 0.00650 mol

2) CH_{3}NH_{2} and CH_{3}NH_{3}^{+} that are in solution:

CH_{3}NH_{2}---> 0.0200 mol − 0.00650 mol = 0.0135 mol

CH_{3}NH_{3}^{+}---> 0.00650 mol

3) The Henderson-Hasselbalch:

pH = 10.643 + log (0.0135 / 0.00650)pH = 10.643 + log 2.076923

pH = 10.32

Note that from (a) to (b) and then from (b) to (c), that pH is becoming more acidic. It's moving from 11.97 to 11.60 to 10.32. Since we are adding an acid (HCl), the pH is doing what it's supposed to do.

Yay!

**Solution to (d):**

At the equivalence point, all the base (the methyl amine) has been converted to its salt, the methylammonium ion. Being the salt of a weak base, the methylammonium ion will hydrolyze, forming an acidic pH at the equivalence point.

1) This is the reaction to consider:

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

2) We can write a K_{a} expression . . .

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

3) . . . and solve it:

(x) (x) 2.2727 x 10¯ ^{11}=–––––––– 0.06667 x = 0.0000012309 M <--- [H

_{3}O^{+}]

4) the pH:

pH = −log 0.0000012309 = 5.91

Notice the concentration of 0.06667 M. That is the [CH_{3}NH_{3}^{+}]. There were 0.0200 moles of it (from all the CH_{3}NH_{2} being neutralized and it was in 300.0 mL total solution. The other 200.0 mL came from the HCl solution required to neutralize all the methyl amine.

Make sure you recognize when you have to have a molarity in the calculation (when doing K_{a} or K_{b} calculations) and when you can use moles or molarity (when doing a Henderson-Hasselbalch calculation).

**Solution to (e):**

The HCl is now in excess. We need to determine how much HCl remains as some of it reacts with the CH_{3}NH_{2}. We will completely ignore the presence of the CH_{3}NH_{3}^{+}, a weak acid, because its influence on the pH is completely overwhelmed by the presence of the HCl, a strong acid.

1) CH_{3}NH_{2} present:

(0.200 mol/L) (0.1000 L) = 0.0200 mol

2) HCl present:

(0.100 mol/L) (0.3000 L) = 0.0300 mol

3) HCl present after reaction with the methyl amine:

0.0300 − 0.0200 = 0.0100 mol

4) New molarity of the HCl:

0.0100 mol / 0.4000 L = 0.025 M

5) pH of the HCl solution:

pH = −log 0.025 = 1.60

**Solution to (f):**

1) This is the chemical equation of interest:

CH_{3}NH_{2}+ H_{3}O^{+}⇌ CH_{3}NH_{3}^{+}+ H_{2}OSome HCl was added, such that we wind up with a buffer of pH 10.643. By the way, we know it is a buffer because the pH at the equivalence point is 5.91. We are well away from that 5.91 value.

2) That means we can write a Henderson-Hasselbalch Equation for the above reaction:

[CH _{3}NH_{2}]pH = pK _{a}+ log–––––––––– [CH _{3}NH_{3}^{+}]

3) The usual situation is that we are solving for the pH. However, in this situation, the pH is known:

[CH _{3}NH_{2}]10.643 = 10.643 + log –––––––––– [CH _{3}NH_{3}^{+}]

4) Since 10.643 − 10.643 is zero, we have:

[CH _{3}NH_{2}]log –––––––––– = 0 [CH _{3}NH_{3}^{+}]

5) Antilog:

[CH _{3}NH_{2}]–––––––––– = 1 [CH _{3}NH_{3}^{+}]

6) The key point now is that the ratio of base to acid is 1:1 at a pH of 10.643. That means that exactly half the base was neutralized by HCl. The will give us the volume of HCl required:

moles CH_{3}NH_{2}originally present(0.200 mol/L) (0.1000 L) = 0.0200 molexactly 0.0100 mol of CH

_{3}NH_{2}was neutralized. (Remember, half the CH_{3}NH_{2}gets neutralized.)this requires 0.0100 mol of HCl

what volume of 0.100 M HCl contains 0.0100 mol?

0.0100 mol / 0.100 mol/L = 0.100 L = 100. mL

Often, in problems of this nature, one of the question parts asks for the pH at the half-equivalence point. in this case, the pH of the half-equivalence point was given and the amount of acid required to get there was asked for.

**Example #3:** Calculate the pH, when the following solutions are added to 100. mL 0.10 M HClO solution:

(a) 0 mL 0.10 M NaOH

(b) 75.0 mL 0.10 M NaOH

(c) 100. mL 0.10 M NaOH

The K_{a} of HClO is 3.0 x 10¯^{8}

**Solution to (a):**

This is a standard weak acid calculation.K

_{a}= ([H^{+}] [ClO¯]) / [HClO]3.0 x 10¯

^{8}= [(x) (x)] / 0.10x = 5.477 x 10¯

^{5}MpH = −log 5.477 x 10¯

^{5}= 4.26

**Solution to (b):**

This is a buffer calculation requiring the use of the Henderson-Hasselbalch Equation.moles HClO ---> (0.10 mol/L) (0.100 L) = 0.010 mol

moles NaOH ---> (0.10 mol/L) (0.0750 L) = 0.00750 molHClO and NaOH react in a 1:1 molar ratio. The HClO is in excess.

moles HClO remaining ---> 0.010 mol − 0.00750 mol = 0.0025 mol

moles NaClO produced ---> 0.00750 molThis problem can be solved by using the moles directly, since they are in the same ratio as the molarities would be.

pH = pK

_{a}+ log [salt / acid]pH = 7.523 + log (0.00750 / 0.0025)

pH = 8.00

**Solution to (c):**

1) It can be seen by examination that equimolar amounts of HClO and NaOH are present. This means that only NaClO is present after reaction. The problem becomes to determine the pH of the salt of an acid knowing the K_{a} of the acid. The first thing to be determine is the K_{b} of the salt.

K_{a}K_{b}= K_{w}(3.0 x 10¯

^{8}) (K_{b}) = 1.0 x 10¯^{14}K

_{b}= 3.333 x 10¯^{7}

2) We need to know the molarity of the NaClO:

0.010 mol / 0.200 L = 0.050 M

3) Use the K_{b} expression:

K_{b}= ([HClO] [OH¯]) / [ClO¯]3.333 x 10¯

^{7}= [(x) (x)] / 0.050x = 0.0001291 M

4) Since this is a hydroxide concentration, we determine the pOH and use that and pK_{w} to determine the pH:

pOH =−log 0.0001291 = 3.89pH = 14.00 − 3.89 = 10.11

**Example #4:** 500. mL of a solution containing 1.50 M NH_{3}(aq) is mixed with 500. mL of a solution containing 0.500 M of HCl(aq). (K_{b} for NH_{3} = 1.77 x 10¯^{5}). Dtermine the pH of:

(a) the final solution.

(b) the solution in (a) after 40.0 mL of 0.200 M NaOH has been added.

(c) the solution in (a) after 24.0 mL of 0.500 M HCl has been added.

**Solution to (a):**

1) The chemical reaction of interest is this (written in net-ionic form):

NH_{3}+ H^{+}---> NH_{4}^{+}

2) It should be fairly obvious that the H^{+} is the limiting reagent. However, let us check:

NH_{3}---> (1.50 mol/L) (0.500 L) = 0.750 mol

H^{+}---> (0.500 mol/L) (0.500 L) = 0.250 molSince NH

_{3}and H^{+}react in a 1:1 molar ratio, it can be seen that H^{+}is the limiting reagent.We can also intuit that H

^{+}is the limiting reagent from the problem type. If the H^{+}is in excess, we wind up with a solution of a strong acid. That's not the usual place to start in a buffer calculation. Moving on . . .

3) Determine moles of NH_{3} remaining and moles of NH_{4}^{+} formed:

NH_{3}---> 0.750 − 0.250 = 0.500 mol

NH_{4}^{+}---> 0.250 mol

4) Use the K_{b} of ammonia to get the K_{a} of ammonium ion:

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

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

_{a}= 5.65 x 10¯^{10}<--- use this to get the pK_{a}

5) Unleash the Henderson-Hasselbalch:

pH = pK_{a}+ log [base / acid]pH = 9.248 + log [0.500 / 0.250)

pH = 9.248 + 0.301 = 9.549

**Solution to (b):**

1) Let us determine how many moles of NaOH are added:

(0.200 mol/L) (0.0400 L) = 0.00800 mol

2) The hydroxide will react with NH_{4}^{+} to form more NH_{3}. The NH_{4}^{+} amount will go down.

NH_{3}---> 0.500 + 0.00800 = 0.508 mol

NH_{4}^{+}---> 0.250 − 0.00800 = 0.242 mol

3) Unleash the Henderson-Hasselbalch:

pH = pK_{a}+ log [base / acid]pH = 9.248 + log [0.508 / 0.242)

pH = 9.248 + 0.322 = 9.570

**Solution to (c):**

1) Let us determine how many moles of HCl are added:

(0.500 mol/L) (0.0240 L) = 0.0120 mol

2) The hydrogen ion will react with NH_{3} to form more NH_{4}^{+}. The NH_{3} amount will go down.

NH_{3}---> 0.500 − 0.0120 = 0.488 mol

NH_{4}^{+}---> 0.250 + 0.0120 = 0.262 mol

3) Unleash the Henderson-Hasselbalch:

pH = pK_{a}+ log [base / acid]pH = 9.248 + log [0.488 / 0.262)

pH = 9.248 + 0.270 = 9.518

**Example #5:** 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) + H_{2}O(ℓ) ⇌ C_{6}H_{5}NH_{2}(aq) + H_{3}O^{+}(aq)

(x) (x) 2.4 x 10¯ ^{5}=––––– 0.100 x = 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:

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

[base] pH = pK _{a}+ log––––– [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:

0.00370 pH = 4.620 + log ––––––– 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

**Example #6:** Consider the titration of 20.0 mL of 0.243 M of KX with 0.106 M HCl. The pK_{a} of HX is 10.39. Give all pH values to 0.01 pH units.

(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 27.9 mL of acid?

(e) What is the pH of the solution after the addition of 55.0 mL of acid?

**Solution to (a):**

1) KX is the salt of a weak acid. It hydrolyzes as follows:

X¯ + H_{2}O ⇌ HX + OH¯

2) X¯ is a base, so we write the K_{b} expression:

[HX] [OH¯] K _{b}=––––––––––– [X¯]

3) The K_{b} is determined from the pK_{a}:

pK_{a}= 10.39, therefore pK_{b}= 3.61K

_{b}= 10¯^{pKb}= 10¯^{3.61}= 2.4547 x 10¯^{4}

4) Solve for the hydroxide concentration:

(x) (x) 2.4547 x 10¯ ^{4}=––––– 0.243 x = 0.010512 M <--- kept a couple guard digits

5) The pOH, then the pH:

pOH = −log 0.010512 = 1.98pH = 12.02

**Solution to (b):**

1) KX and HCl react in a 1:1 molar ratio:

KX + HCl ---> HX + KCl

2) Moles of KX that got titrated:

(0.243 mol/L) (0.0200 L) = 0.00486 mol

3) How many mL of 0.106 M HCl are required to deliver 0.00486 mol?

0.00486 mol / 0.106 mol/L = 0.04585 L = 45.85 mL

I decided to ignore the solution path that uses millimoles.

**Solution to (c):**

N.B. The solution at the equivalence point is the solution of a weak acid (HX). The KCl plays no role in the pH calculation.

1) Calculate the molarity of HX in the solution at the equivalence point:

0.00486 mol / 0.06585 L = 0.0738 MThe 0.06585 L came from the 45.85 mL of HCl solution and the 20.0 mL of KX solution.

2) Write the K_{a} expression for HX and solve:

HX + H_{2}O ⇌ H_{3}O^{+}+ X¯

[H _{3}O^{+}] [X¯]K _{a}=––––––––––– [HX]

(x) (x) 4.0738 x 10¯ ^{11}=–––––– 0.0738 N.B. the K

_{a}came from 10¯^{pKa}= 10¯^{10.39}x = 0.0000235 M

pH = −log 0.0000235 = 4.63

**Solution to (d):**

At this point the titration is only partially done (remember, it required 45.85 mL of HCl to reach the equivalence point). The 27.9 mL in this part of the problem will create a buffer and we will use the Henderson-Hasselbalch equation to get the pH.

1) Determine moles of HCl added:

(0.106 mol/L) (0.0279 L) = 0.0029574 mol

2) The HCl was added to 0.00486 mol of KX. How much HX and X¯ are in the solution?

moles HX = 0.0029574 mol (all the HCl reacted)

moles X¯ remaining ---> 0.00486 mol − 0.0029574 mol = 0.0019026 mol

3) Unleash the Henderson-Hasselbalch!

0.0019026 pH = 10.39 + log ––––––––– 0.0029574 N.B. I didn't have to calculate molarities because the ratio of the moles involved is equal to the ratio of molarities.

pH = 10.39 + (−0.19)

pH = 10.20

This pH fits. We are titrating a base (X¯) with an acid (HCl). The pH has moved from 12 to 10 on its way to 4.6.

**Solution to (e):**

The solution has been titrated past the endpoint and there is now excess strong acid in the solution. This overwhelms the presence of the weak acid (the HX) and the solution acts like a solution with only HCl present.

1) Calculate the volume of unreacted HCl:

55.0 mL − 45.85 mL = 9.15 mL

2) Calculate the new molarity of the 9.15 mL of HCl:

M_{1}V_{1}= M_{2}V_{2}(0.106 mol/L) (9.15 mL) = (x) (75.0 mL)

x = 0.012932 M

N.B. The 75 mL comes from 20 mL of KX being titrated with 55 mL of acid.

3) Calculate the pH:

pH = −log 0.012932 = 1.89N.B. HCl is a strong acid, so it dissociates 100%

**Bonus Example:** 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}:

0.001103 3.72 = pK _{a}+ log––––––––––– 0.002031726 3.72 = pK

_{a}+ (−0.265)pK

_{a}= 3.98 (to two, not three, sig figs)10¯

^{pKa}yields the K_{a}