When a very strong aqueous acid reacts with a very strong aqueous base, the net reaction occurs between the hydronium and hydroxide ions.

Transcription

1 Experiment 2-A ACID-BASE TITRATIONS CHM 1041 The Equivalent Mass of an Unknown Acid EqWt.wpd DISCUSSION Brknsted-Lowry theory defines an acid as a hydrogen ion (H + ) donor and a base as a hydrogen ion (H + ) acceptor. Some compounds are amphiprotic, i.e., they have the ability to act as an acid or a base. Water has this potential: H 2 O + H + H 3 O + H 2 O OH - + H + base acid In aqueous solutions, very strong acids force water to act as a base and one hundred percent of the acidic molecules ionize. For example, when molecular hydrogen chloride gas is bubbled through water it ionizes completely to yield hydronium ions and chloride ions in solution: HCl (g) + H 2 O (l) H 3 O + (aq) + Cl - (aq) 100% ionization Therefore in water solutions, acids much stronger than H 3 O + are completely converted to H 3 O +, and it follows that the strongest acid possible in water 1 in appreciable concentration is the hydronium ion, H 3 O +. Very strong bases cause water to act as an acid, thereby generating the hydroxide ion, OH -. For example, oxide ion reacts completely with water giving only hydroxide ion. O 2- + H 2 O OH - + OH - Bases much stronger than OH - are completely converted to OH - when placed in water solvent, and therefore the strongest base that can exist in water 2 in appreciable concentration is OH -. Many ionic oxides react with water to give hydroxide solutions, for example Na 2 O (s) + H 2 O (o) 2 Na + (aq) + 2 OH - (aq) When a very strong aqueous acid reacts with a very strong aqueous base, the net reaction occurs between the hydronium and hydroxide ions. H 3 O + (aq) + OH - (aq) 2 H 2 O (o) This is the actual reaction that occurs when a very strong aqueous acid is neutralized with a very strong aqueous base. Note that one mole of hydronium ions reacts exactly with one mole of hydroxide ions to accomplish neutralization, and one mole of hydrogen ions (H + ) has been transferred from acid to base. Acids weaker than hydronium ion only partially ionize in water solution. An equilibrium is 1 H3 O + is the conjugate acid of water. In nonaqueous solvents, the strongest acid possible would be the conjugate acid of the solvent, e.g., in liquid methanol, CH 3 OH + H + + CH 3 OH 2 2 OH - is the conjugate base of water. In nonaqueous solvents, the strongest base possible would be the conjugate base of the solvent, e.g., in liquid methanol, CH 3 OH - H + CH 3 O -

2 2 established in which the weaker molecular acid (reactant) is the acid present in solution in greater abundance. Consider 0.1 M HCO 2 H (formic acid): HCO 2 H (aq) + H 2 O (aq) H 3 O + (aq) + HCO 2 -(aq) J4% ionization Neutralization of an aqueous solution of a moderate acid and a very strong base therefore would be best described as a reaction between the molecular acid and the hydroxide ion. For example, consider the reaction between formic acid and hydroxide ion: HCO 2 H (aq) + OH - (aq) H 2 O (o) + HCO 2 -(aq) Here again, one mole of hydrogen ions has been transferred from the acid to the base. Formic acid has only one acidic hydrogen per molecule and is therefore described as monoprotic. Many acids, however, have more than one acidic hydrogen per molecule and are capable of donating more than one mole of hydrogen ions per mole of acid. For example, oxalic acid is diprotic since it contains two acidic hydrogens per molecule, and requires two moles of hydroxide base per mole of acid to accomplish neutralization. H 2 C 2 O 4 (aq) + 2 OH - (aq) C 2 O 4 2-(aq) + 2 H 2 O (o) The point at which an acid-base reaction is stoichiometrically complete may be referred to as the neutralization or equivalence point. At this condition, the hydrogen ion donating capacity of the acid has just been matched by the hydrogen ion accepting capacity of the base. Equivalent Mass and Normality The equivalent mass (or equivalent weight) of an acid or base is the mass of the acid or base that donates or accepts one mole of hydrogen ions in the reaction being considered. One equivalent mass is also called one equivalent (1 eq). As an example, consider the following acid-base reaction: H 3 PO 4 (aq) + 3 OH - (aq) PO 4 3-(aq) + 3 H 2 O (o) One mole of phosphoric acid donates three moles of hydrogen ions in this reaction and is behaving as a triprotic acid; 3 therefore 1 mole = 3 eq. The equivalent mass of H 3 PO 4 may now be calculated: 98.0 g x 1 mol = 32.7 g mol 3 eq eq This means that 32.7 grams of H 3 PO 4 donate one mole of hydrogen ions when it reacts with enough hydroxide. The equivalent mass is always related to the molar mass by a small whole number (1,2, or 3), this number being the number of hydrogen ions donated/accepted by one mole of the acid/base that is reacting. Some other examples of acids/bases and their common equivalent relationships follow. 3 H3 PO 4 may behave as a diprotic acid with bases weaker than OH -, e.g., ammonia. H 3 PO 4 (aq) + 2 NH 3 (aq) 2 NH 4 + (aq) + HPO 4 2-(aq) In this reaction, 1 mole of H 3 PO 4 equals 2 equivalents.

3 3 Ca(OH) 2 dibasic, therefore... 1 mole = 2 eq 74.1 g x 1 mol = 37.0 g mol 2 eq eq H 2 SO 4 diprotic, therefore... 1 mole = 2 eq 98.1 g x 1 mol = 49.0 g mol 2 eq eq HC 2 H 3 O 2 monoprotic, therefore... 1 mole = 1 eq 60.1 g x 1 mol = 60.1 g mol 1 eq eq A solution's concentration is often conveniently expressed as the molarity of that solution, M. The molarity is the number of moles of the solute dissolved in one liter of the solution (moles/l). An alternative for expressing a solution's concentration that is based on the number of equivalents of solute rather than the number of moles is called normality, N. The normality is the number of equivalents of the solute dissolved in a liter of solution (eq/l). Molarity and normality are easily related by small whole numbers: 0.25 M H 3 PO 4 triprotic, therefore... 1 mole = 3 eq 0.25 mol x 3 eq = 0.75 eq or 0.75 N H 3 PO 4 L mol L 4.82 x 10-3 N Ba(OH) 2 dibasic, therefore... 1 mole = 2 eq 4.82 x 10-3 eq x 1 mol = 2.41 x 10-3 mol or 2.41 x 10-3 M L 2 eq L Volumetric Analysis Analysis of an unknown sample of an acid may be accomplished by reaction with a known sample of a base to the point of equivalence (or an unknown base may be analyzed with a known acid). The fundamental premise here is that the number of equivalents of base will be equal to the number of equivalents of acid at neutralization. Titration is the procedure used to accurately measure the volume of a reactant solution which is stoichiometrically equivalent to an accurately measured quantity of a second reactant. The reactant solution is called the titrant and is delivered by a buret. An indicator, a substance which changes color, is added to detect the equivalence or end-point. If the normality of the titrant is accurately known, the number of equivalents of titrant may be calculated: N x V = Number of equivalents If the titrant was a basic solution of known normality, the number of equivalents of base delivered may be calculated, and this is equal to the number of equivalents of unknown acid that reacted with the base. For example, how many equivalents of unknown acid are present if the acid was titrated to the equivalence point with ml of N KOH solution?

4 4 N x V = eq x L = eq or x 10-3 eq KOH L Therefore, x 10-3 equivalent of unknown acid was neutralized by the KOH titrant. EXPERIMENTAL In part A, you will determine the normal concentration (standardize) of a stock solution of sodium hydroxide. In part B, you will use the standard sodium hydroxide solution to titrate weighed samples of an unknown acid, and then calculate the equivalent mass of the unknown acid. A. The Standardization of a Sodium Hydroxide Solution A stock solution of unknown concentration has been prepared for use as the titrant in part B of this experiment. First the normality of the of the NaOH titrant must be accurately determined. This can be done by titrating the NaOH solution with a solution of a solid acid that can be accurately weighed. 4 The solid acid, known as a primary standard, must be of high purity and remain free of moisture. The primary standard used in this experiment is potassium hydrogen phthalate, KHC 8 H 4 O 4, which is monoprotic and abbreviated "KHP". The KHP is dissolved in water and titrated with the NaOH to the equivalence point using phenolphthalein indicator. 5 Three standardization titrations of KHP with the stock NaOH solution have been done for you. Obtain these data and carefully record it on the following data table. You will use these data to calculate the normal concentration of the stock NaOH solution. In the space provided, show the method/set-up for each set of calculations for each titration. INCLUDE UNITS. Finally, calculate the average normality for the stock NaOH solution. This is the normality value that will be used in part B of this experimental section. Standardization of Stock NaOH Solution Include correct units and significant figures Note: KHP L KHC 8 H 4 O 4, Molar Mass KHC 8 H 4 O 4 4 The stock NaOH solution has been prepared by dissolving about 4 grams of NaOH in a liter of boiled, cooled water. The water was boiled to expel dissolved carbon dioxide, which reacts with water to form carbonic acid, H 2 CO 3. The carbonic acid would react with and consume hydroxide ion to yield carbonate ion. CO 2 (g) + H 2 O (l) H 2 CO 3 (aq) H 2 CO 3 (aq) + 2 OH - (aq) CO 3 2-(aq) + 2 H 2 O (l) Also, NaOH cannot be used as a primary standard because of its tendency to rapidly absorb moisture. An accurate weighing of pure NaOH is impractical. 5 In this experiment the indicator used to determine the end-point of the titration is called phenolphthalein, which is abbreviated HPth. The HPth shows a different color in acid solution than in base solution, where it becomes the ion Pth -. HPth (aq) + OH - (aq) H 2 O (l) + Pth - (aq) colorless pink HPth is a weaker acid than any of the acids titrated in this experiment. When all of the titrated acid has been consumed (equivalence point), the OH - starts to react with the HPth, forming Pth -, and the color of the solution suddenly changes from colorless to pink, marking the end-point of the titration.

5 5 Run 1 Run 2 Run 3 Mass of KHP Moles of KHP Since KHP is monoprotic, 1 mole KHP = 1 equivalent KHP Equivalents KHP Final buret reading Initial buret reading Total Volume NaOH Titrated, ml Volume NaOH, L Normality NaOH Average Normality NaOH Solution Calculations (Include set-up and units): Titrations with NaOH Solution B. The Equivalent Mass of an Unknown Acid The equivalent mass of an unknown solid acid will be determined as follows. An accurately measured mass of solid acid will be titrated to the equivalence point with the standardized NaOH solution. From the volume of the titrated NaOH solution and its known normality is found the number of equivalents of NaOH delivered. The number of equivalents of NaOH should be equal to the number of equivalents of unknown acid. Having found the number of equivalents of acid in

6 6 the sample and its accurate mass in grams, the equivalent mass of the acid (g/eq) may be calculated. Experimental Procedure 1) Obtain a 50-mL buret, four clean and dry 125-mL Erlenmeyer flasks for sample titrations, and a wash bottle filled with deionized water. 2) Next, obtain about 100 ml of stock NaOH solution and save it in a clean, dry, labeled Erlenmeyer flask. Keep the NaOH solution stoppered to minimize reaction with CO 2 from the air. 3) Rinse the inside of the buret with hot water and allow it to drain through the stopcock and tip. Rinse it again, this time with deionized water, and allow the buret to drain upside down affixed to the buret clamp. 4) Obtain your sample of unknown solid acid and immediately record its ID number on the data table. 5) Assemble the four, clean, dry, and labeled 125 ml Erlenmeyer flasks. Tare each flask on the analytical balance and then carefully add the approximate mass of acid as indicated on the unknown label. Record on your data table the accurate mass of each sample to the limit of the balance sensitivity, and then cork each flask. 6) Now return to the buret and rinse the inside of the buret twice with about 3 ml of the stock NaOH solution, draining through the stopcock and tip. Affix the buret to the buret clamp right side up, making sure the buret is in a stable, vertical position, and then fill it to just above the 0.00 ml scale division with the NaOH standard solution. Open the stopcock briefly to fill the tip with liquid, taking care to expel air bubbles. The liquid level should now fall below the 0.00 ml mark. Allow the liquid a few moments to settle. 7) Add 25 ml of deionized water and 3 drops of phenolphthalein to each of the Erlenmeyer flasks containing your unknown solid acids. Swirl the flasks to dissolve the solids. 8) Locate the first sample flask under the buret with the tip well inside the mouth of the flask. Place a piece of white paper under the flask. Record the initial buret reading on the data table. Remember to estimate buret readings to the nearest 0.01 ml. Do not attempt to adjust the initial reading to the 0.00 ml mark. 9) Add the NaOH solution from the buret to the sample in the flask while swirling the flask. At some time during the titration you will see a flash of pink color where the stream of NaOH mixes with the solution in the flask. When this pink color begins to fade more slowly, slow the addition of the NaOH solution from the buret. Ideally, the NaOH solution should be added one drop at a time as the end-point is approached so that the titration can be stopped when one drop of NaOH solution turns the solution in the flask from colorless to a permanent but very faint pink. Just before the end-point is reached, rinse the inside of the flask with the deionized water from your wash bottle. Record the final buret reading at the end-point on the data table. Repeat this procedure with the remaining samples and check your precision when four runs are completed. A "Q Test" may be applied to eliminate the least precise run. The Equivalent Mass of an Unknown Acid Unknown Acid ID Number Normality Std NaOH Soln

7 7 Run 1 Run 2 Run 3 Run 4 Mass of unknown acid Final buret reading Initial buret reading Volume NaOH Titrated, ml Precision ml of NaOH Check g unknown Volume NaOH, L Number Equiv. of NaOH Number Equiv. of Acid Equivalent Mass of Acid Average Equivalent Mass of Acid Calculations (Include set-up and units): Questions 1. Given the following reagent equation for the standardization of the stock NaOH solution with potassium hydrogen phthalate, KHC 8 H 4 O 4, write the net equation. KHC 8 H 4 O 4 (aq) + NaOH (aq) NaKC 8 H 4 O 4 (aq) + H 2 O (l)

8 2. Given the Lewis structure for KHP, KHC 8 H 4 O 4, explain the fact that KHP is quite soluble in water. 3. The inventory of aqueous 1 M H 2 SO 4 is: [H 3 O + ] = 1 M and [HSO 4 - ] = 1 M. Given this information, write the correct equation(s) for any Brknsted-Lowry acid/base reaction(s) which occur... a) when 1 ml of 1M H 2 SO 4 is mixed with 1 drop of 1 M NaOH. b) when 1 drop of 1 M H 2 SO 4 is mixed with 1 ml of 1 M NaOH DATA/REPORT Experiment 2A, Acid-Base Titrations Standardization of Stock NaOH Solution Include correct units and significant figures Name Section Date Note: KHP L KHC 8 H 4 O 4, Molar Mass KHC 8 H 4 O 4 Run 1 Run 2 Run 3

9 9 Mass of KHP Moles of KHP Since KHP is monoprotic, 1 mole KHP = 1 equivalent KHP Equivalents KHP Final buret reading Initial buret reading Total Volume NaOH Titrated, ml Volume NaOH, L Normality NaOH Average Normality NaOH Solution Calculations (Include set-up and units): Titrations with NaOH Solution Equivalent Mass of an Unknown Acid Unknown Acid ID Number Normality Std NaOH Soln Run 1 Run 2 Run 3 Run 4

10 10 Mass of unknown acid Final buret reading Initial buret reading Volume NaOH Titrated, ml Precision ml of NaOH Check g unknown Volume NaOH, L Number Equiv. of NaOH Number Equiv. of Acid Equivalent Mass of Acid Average Equivalent Mass of Acid Calculations (Include set-up and units): Questions 1. Given the following reagent equation for the standardization of the stock NaOH solution with potassium hydrogen phthalate, KHC 8 H 4 O 4, write the net equation. KHC 8 H 4 O 4 (aq) + NaOH (aq) NaKC 8 H 4 O 4 (aq) + H 2 O (o)

11 11 3. Given the Lewis structure for KHP, KHC 8 H 4 O 4, explain the fact that KHP is quite soluble in water. 3. The inventory of aqueous 1 M H 2 SO 4 is: [H 3 O + ] = 1 M and [HSO 4 - ] = 1 M. Given this information, write the correct equation(s) for any Brknsted-Lowry acid/base reaction(s) which occur... a) when 1 ml of 1M H 2 SO 4 is mixed with 1 drop of 1 M NaOH. b) when 1 drop of 1 M H 2 SO 4 is mixed with 1 ml of 1 M NaOH 4. SAVE your unknown number and the equivalent mass of your unknown acid as information to be applied in an upcoming experiment (Experiment 3, Freezing Point Depression, Molar Mass Determination in t-butyl Alcohol).