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Titration of Acids and Bases

Let's become familiar with the techniques of titration; to determine the molecular weight of a solid acid.

Equipment: 2 50-mL burets, 1000-mL Erlenmeyer flask, 3 250-mL Erlenmeyer flasks, weighing paper, ring stand, ring, wire gauze, analytical balance, Bunsen burner, rubber hose, 1-quart bottle with screw top, squirt bottle, buret clamp, stopcock, microspatula, small beaker

Materials: 19 M NaOH, potassium hydrogen phthalate (KHP), phenolphthalein solution, citric acid (H3C6H5O7), distilled water, petroleum jelly, unknown HCl solution

Introduction: One of the most common and familiar reactions in chemistry is the reaction of an acid with a base. This reaction is called neutralization, and the essential feature of this process in aqueous solution is the combination of hydronium ions with hydroxide ions to form water. In this experiment, this reaction will be used to determine accurately the concentration of a sodium hydroxide solution. The process of determining the concentration of a solution is called standardization.

The technique of accurately measuring the volume of a solution required to react with another reagent is termed titration. The titration process is shown in Figure 20.2 below:

An indicator solution is used to determine exactly when an acid has neutralized a base or vice versa. An indicator changes colors when equivalent amounts of acid and base are present. The color change is termed the “end point” of the titration. Indicators change colors at different pH values. Phenolphthalein, the indicator used in this experiment, changes from colorless to pink at a pH of 9. In slightly more acidic solutions it is colorless, and in more alkaline solutions it is pink.

In this experiment, sodium hydroxide will be titrated against three solutions: KHP, citric acid, and hydrochloric acid. KHP, or potassium hydrogen phthalate, is a molecule with the following structural diagram:

The goal of this experiment is to standardize sodium hydroxide against KHP; to determine the molecular weight of citric acid; and to determine the concentration of the unknown hydrochloric acid. In order to determine the concentration of the hydrochloric acid, the formula must be as follows: molarity = moles solute/volume of solution in liters. 

Procedure:

A. Standardization of NaOH solution

1. Carbon-dioxide free water was prepared by boiling tap water.
2. Three samples of pure KHP were weighed with weighing paper. Between .4 and .6 grams of KHP were obtained per sample. Each KHP sample was placed into a separate
250-mL Erlenmeyer flask.
3. Between 15 and 50 mL of the boiled water was added to each Erlenmeyer flask. The contents of each flask were swirled. Two to four drops of phenolphthalein solution were added to each flask.
4. Some NaOH solution was poured into the buret and allowed to flow through the buret. It was collected in a small beaker. The stopcock was then turned to the closed position. The buret was then completely filled to the 0-mL mark with NaOH solution. The initial buret reading was recorded.
5. NaOH solution was slowly added to one of the flasks containing KHP solution while gently swirling the contents of the flask, as indicated in Figure 20.2 in the Introduction. NaOH solution was added until the KHP turned completely pink. The buret reading was recorded.
6. Steps 4 and 5 were repeated for the two other Erlenmeyer flasks.

B. Analysis of citric acid

1. Three samples of pure citric acid were weighed with weighing paper. Between .2 and
.3 grams of citric acid were obtained per sample. Each citric acid sample was placed into a separate 250-mL Erlenmeyer flask.
2. Between 15 and 50 mL of the previously boiled water was added to each Erlenmeyer flask. The contents of each flask were swirled. Two to four drops of phenolphthalein solution were added to each flask.
3. Some NaOH solution was poured into the buret and allowed to flow through the buret. It was collected in the same small beaker used in step 4 of part A. The stopcock was then turned to the closed position. The buret was then completely filled to the 0-mL mark with NaOH solution. The initial buret reading was recorded.
4. NaOH solution was slowly added to one of the flasks containing citric acid solution while gently swirling the contents of the flask, as indicated in Figure 20.2 in the Introduction. NaOH solution was added until the citric acid turned completely pink. The buret reading was recorded.
5. Steps 3 and 4 were repeated for the two other Erlenmeyer flasks.

C. Analysis of unknown hydrochloric acid solution

1. An arbitrary amount of HCl was placed into an Erlenmeyer flask from a second, clean buret. Twice as much HCl was placed into another Erlenmeyer flask; three times as much as the original amount of HCl was placed into a third Erlenmeyer flask. A few drops of phenolphthalein solution were added to each flask. Some distilled water was added to each flask for better visibility.

2. Some NaOH solution was poured into the original buret and allowed to flow through the buret. It was collected in the same small beaker used in step 4 of part A. The stopcock was then turned to the closed position. The buret was then completely filled to the 0-mL mark with NaOH solution. The initial buret reading was recorded.

3. NaOH solution was slowly added to the first Erlenmeyer flask containing HCl. The contents of the flask were gently swirled during this titration. NaOH solution was added until the HCl turned completely pink. The buret reading was recorded.

4. Steps 2 and 3 were repeated for the two other Erlenmeyer flasks.

Observations:

± will not be factored into results for part A. For mass observations, ± is .0001 g. For buret readings, ± is .05 mL.

A. Standardization of NaOH solution

B. Analysis of citric acid

C. Analysis of unknown hydrochloric acid solution

Results:

A. Standardization of NaOH solution

mL of NaOH used (final buret reading minus initial buret reading) = Trial 1. 20.90 mL – 0.00 mL = 20.90 mL.
Trial 2. 23.80 mL – 0.00 mL = 23.80 mL. Trial 3. 22.90 mL – 0.00 mL = 22.90 mL.
Moles of KHP used (mass of KHP used times (1 mole KHP/204.2 grams KHP)) = Trial 1. (.4730 g KHP (1 mole/204.2 g)) = .002316 moles KHP.
Trial 2. (.5311 g KHP (1mole/204.2 g)) = .002601 moles KHP. Trial 3. (.5159 g KHP (1 mole/204.2 g)) = .002526 moles KHP.
Moles of NaOH used (moles of KHP used times (1 mole NaOH/1 mole KHP)) = Trial 1. (.002316 moles KHP (1 mole NaOH/1 mole KHP)) = .002316 moles NaOH. Trial 2. (.002601 moles KHP (1 mole NaOH/1 mole KHP)) = .002601 moles NaOH. Trial 3. (.002526 moles KHP (1 mole NaOH/1 mole KHP)) = .002526 moles NaOH. Molarity of NaOH (moles of NaOH divided by (mL of NaOH used/1000)) =
Trial 1. .002316 moles NaOH / .02090 L = .1108 M Trial 2. .002601 moles NaOH / .02380 L = .1093 M Trial 3. .002526 moles NaOH / .02290 L = .1103 M Average molarity (sum of molarities divided by 3) = (.1108 + .1093 + .1103) / 3 = .1101 M
Standard deviation (|Average molarity minus individual molarities|) = Trial 1. |.1101 M .1108 M| = .0007 M
Trial 2. |.1101 M .1093 M| = .0008 M
Trial 3. |.1101 M .1103 M| = .0002 M
Average deviation (Sum of standard deviations divided by 3) = (.0007 M + .0008 M + .0002 M) / 3 = .0006 M

B. Analysis of citric acid

mL of NaOH used (final buret reading minus initial buret reading) = Trial 1. 34.50 mL – 0.00 mL = 34.50 mL
Trial 2. 36.30 mL – 0.00 mL = 36.30 mL Trial 3. 34.30 mL – 0.00 mL = 34.30 mL
Moles of NaOH used (molarity of NaOH times (mL of NaOH/1000)) = Trial 1. (.1101 ± .0006 M (34.50 ± .05 mL/1000)) =
(.1101 ± .54% M (.03450 ± .14% L)) = (.003798 ± .68% mol) =
.003798 ± .00003 mol = .00380 ± .00003 mol
Trial 2. (.1101 ± .0006 M (36.30 ± .05 mL/1000)) =
(.1101 ± .54% M (.03630 ± .14% L)) = (.003997 ± .68% mol) =
.003997 ± .00003 mol = .00400 ± .00003 mol
Trial 3. (.1101 ± .0006 M (34.30 ± .05 mL/1000)) =
(.1101 ± .54% M (.03430 ± .15% L)) = (.003776 ± .69% mol) =
.003776 ± .00003 mol = .00378 ± .00003 mol
Moles of citric acid used (moles of NaOH used (1 mole citric acid/3 moles NaOH)) = Trial 1. (.00380 ± .00003 mol NaOH (1 mole citric acid/3 moles NaOH)) =
(.00380 ± .79% moles citric acid / 3) = (.00126 ± .79% moles citric acid) =
.00126 ± .00001 moles citric acid
Trial 2. (.00400 ± .00003 mol NaOH (1 mole citric acid/3 moles NaOH)) =
(.00400 ± .75% moles citric acid / 3) = (.00133 ± .75% moles citric acid) =
.00133 ± .00001 moles citric acid
Trial 3. (.00378 ± .00003 mol NaOH (1 mole citric acid/3 moles NaOH)) = (.00378 ± .79% moles citric acid / 3) = (.00126 ± .79% moles citric acid) =
.00126 ± .00001 moles citric acid
Molecular weight of citric acid (grams of citric acid divided by moles of citric acid) = Trial 1. (.2456 g / .00126 mol) = 194.9 g/mol
Trial 2. (.2527 g / .00133 mol) = 190.0 g/mol
Trial 3. (.2407 g / .00126 mol) = 191.0 g/mol
Average molecular weight (sum of molecular weights divided by 3) = (194.9 + 190.0 + 191.0) / 3 = 192.0 g/mol
Standard deviation (|Average molecular weight minus individual molecular weight|) = Trial 1. |192.0 g/mol – 194.9 g/mol| = 2.9 g/mol
Trial 2. |192.0 g/mol – 190.0 g/mol| = 2.0 g/mol
Trial 3. |192.0 g/mol – 191.0 g/mol| = 1.0 g/mol
Average deviation (Sum of standard deviations divided by 3) = (2.9 + 2.0 + 1.0) / 3 = 2.0 g/mol

C. Analysis of unknown hydrochloric acid solution

Amount of NaOH used (final buret reading minus initial buret reading) = Trial 1. 3.65 mL – 0.00 mL = 3.65 mL
Trial 2. 7.30 mL – 0.00 mL = 7.30 mL Trial 3. 10.95 mL – 0.00 mL = 10.95 mL
Moles of NaOH used (molarity of NaOH times (mL of NaOH/1000)) =
Trial 1. (.1101 ± .0006 M (3.65 ± .05 mL/1000)) =
(.1101 ± .54% M (.00365 ± 1.37% L)) = (.000402 ± 1.91% mol) =
.000402 ± .000007 mol
Trial 2. (.1101 ± .0006 M (7.30 ± .05 mL/1000)) =
(.1101 ± .54% M (.00730 ± .68% L)) = (.000804 ± 1.22% mol) =
.00080 ± .00001 mol
Trial 3. (.1101 ± .0006 M (10.95 ± .05 mL/1000)) =
(.1101 ± .54% M (.01095 ± .46% L)) = (.001206 ± 1.00% mol) =
.00121 ± .00001 mol
Moles of HCl used (moles of NaOH (1 mole HCl / 1 mole NaOH) =
Trial 1. .000402 ± .000007 mol (1 mol HCl / 1 mol NaOH) = .000402 ± .000007 mol. Trial 2. .00080 ± .00001 mol (1 mol HCl / 1 mol NaOH) = .00080 ± .00001 mol.
Trial 3. .00121 ± .00001 mol (1 mol HCl / 1 mol NaOH) = .00121 ± .00001 mol. Molarity of unknown HCl solution (moles of HCl used / (Amount of HCl used/1000))= Trial 1. (.000402 ± .000007 mol HCl / (5.00 ± .05 mL / 1000) =
(.000402 ± 1.74% mol HCl / (.00500 ± 1.00% L)) = (.0804 ± 2.74% M) =
.080 ± .002 M
Trial 2. (.00080 ± .00001 mol HCl / (10.00 ± .05 mL / 1000) = (.00080 ± 1.25% mol HCl / (.0100 ± .50% L)) = (.08 ± 1.75% M) =
.080 ± .001 M
Trial 3. (.00121 ± .00001 mol HCl / (15.00 ± .05 mL / 1000) = (.00121 ± .83% mol HCl / (.0150 ± .33% L)) = (.081 ± 1.16% M) =
.081 ± .001 M
Average molarity (sum of molarities divided by 3) = (.080 + .080 + .081) / 3 = .080 M

Discussion: The value obtained for the molecular weight of citric acid was 192.0 g/mol. The true value is 192.1 g/mol. The percent error calculation is as follows:

(observed-true)/true * 100 (192.0-192.1)/192.1 * 100 =
-.1/192.1 * 100 =
-.05% error.

This value is very close. The experiment was very accurate. Some sources of error in this experiment include faulty equipment. If the buret was leaking at all, then the reading would have been inaccurate. If there was any petroleum jelly in the hole of the stopcock, then this would have served as an impurity, distorting volumetric readings. If there were any air bubbles in the buret, then the volume reading would have been too low, because the meniscus would have been pushed upwards.

The theory associated with this experiment is the process of metathesis reactions. Metathesis, or single-replacement, reactions occur in the following form: AX + BY → AY + BX. There are three driving forces for metathesis reactions to occur: the formation of a precipitate, the formation of a weak electrolyte or nonelectrolyte, or the formation of a gas that escapes from solution. Metathesis reactions occur to neutralize acids and bases, as they did in this experiment.

Ramifications of this experiment are both personal and general. In general, a new procedure was learned and practiced extensively. Specific technical applications of this
experiment include titrating to determine the acid or base content of a solution. For example, titration can be used to determine the acid content of beer. If beer is too acidic or too basic, it will not taste the way it should. Thus, titrating a small sample of beer against a base like sodium hydroxide while using an indicator like phenolphthalein will reveal the acid content. This can be done with any solution.

Questions:

1. The balanced chemical equation for the reaction of KHP with NaOH is: KHC8H4O4(aq) + NaOH(aq) → H2O(l) + KNaC8H4O4(aq)

2. If the KHP sample were contaminated with NaCl before standardization, the molarity
reading would have been too low because the number of moles of NaOH measured would have been too low.

3. To find the molarity of the malonic acid solution, first find the number of moles of
NaOH used.
mol NaOH = molarity * liters
mol NaOH = .100 M * .02182 L = .002182 mol NaOH
Then calculate the number of moles of malonic acid needed.
mol malonic acid = mol NaOH (1 mol malonic acid / 2 mol NaOH)
mol malonic acid = .002182 mol NaOH (1/2) = .001091 mol malonic acid
Then calculate the molarity of malonic acid. molarity = moles / liters
molarity = .001091 mol / .01212 L = .0900 M

4. To find the molarity of the sulfuric acid, first find the number of moles of sodium carbonate used.
mol sodium carbonate = .432 g sodium carbonate (1 mol / 106.0 g)
mol sodium carbonate = .00408 mol
Then find how many moles of sulfuric acid were used.
mol sulfuric acid = .00408 mol sodium carbonate (1 mol sulfuric acid / 1 mol sodium carbonate) = .00408 mol sulfuric acid
Then calculate the molarity of the sulfuric acid. molarity = moles / liters
molarity = .00408 mol / .0223 L = .183 M sulfuric acid.

5. To find the molarity of the solution, first determine the number of moles.
.252 g oxalic acid (1 mole / 126.0 g) = .00200 mol oxalic acid
Then calculate the molarity of the solution. molarity = moles / liters
molarity = .00200 mol / .500 L = .00400 M oxalic acid

Conclusion: The experiment was completed with a high degree of success. Titration techniques were learned and practiced, and the molecular weight of a solid acid was determined accurately.