The buret is filled with the naoh titrant

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STANDARIZATION OF A BASE AND TITRATION OF A VINEGAR SOLUTION

ADDITIONAL READING The concepts in this experiment are also discussed in sections 3.6 AND 17.3 of Chemistry and Chemical Reactivity by Kotz, Treichel, Townsend and Treichel, and in sections 4.3b, 17.3a, and 17.3b of Mindtap General Chemistry by Vining, Young, Day, and Botch

ABSTRACT This experiment is divided into two parts. Each student is expected to perform the experiment individually.

In Part A, you will prepare a NaOH titrant solution, then standardize it (determine its exact concentration) using the acid primary standard, potassium hydrogen phthalate, KHC8H4O4, frequently abbreviated as KHP. Note KHP is not a chemical formula.

In Part B you will use your standardized NaOH solution to determine the molar concentration of vinegar (an acetic acid, CH3COOH, solution), and convert this concentration unit to a mass percent concentration unit, and finally compare your measured mass percent concentration to the value reported on the bottle.

BACKGROUND TITRATIONS One of the most useful strategies in analytical chemistry is to use a known reagent (known composition or concentration) as a standard to analyze an unknown substance. A titration is an analytical procedure in which a solution of known concentration, the standard solution, is slowly reacted with a solution of unknown concentration. The concentration of the unknown solution can be easily calculated. Titration is often used to measure the concentration of an acid or base, but it can also be used for any chemical reaction if the stoichiometry is known.

EXPERIMENTS 6 AND 7 ARE BOTH ACID BASE TITRATION EXPERIMENTS, QUITE SIMILAR TO EACH OTHER. THE REASONS FOR DOING TWO TITRATION EXPERIMENTS

A. TO GIVE STUDENTS PLENTY OF OPPORTUNITY BOTH TO PERFECT THEIR TITRATION TECHNIQUE AND TO LEARN TO DO THE CALCULATIONS;

B. TITRATION IS THE MOST IMPORTANT TECHNIQUE LEARNED IN CHEM 1033 LAB.

YOU WILL DO A PRACTICAL LAB EXAM AT THE END OF THE SEMESTER; IT WILL BE A VERY SIMILAR TITRATION.

IT IS IMPORTANT TO REALIZE THAT TITRATION IS AN ACQUIRED SKILL, REQUIRING PRACTICE. MOST STUDENTS ARE NOT PROFICIENT AT FIRST, BUT IF YOU WANT TO BECOME EXPERT AT IT, YOU WILL GET THERE WITH PRACTICE.

It is critical that there be an observable change that signals that the titration is complete. This is called the endpoint, since it signals the end of the titration, when the equivalents of titrant added just equal the equivalents of the analyte unknown. When performing an acid-base titration, we commonly use an acid-base indicator that has one color before the endpoint but changes sharply to a different color at the pH of the endpoint.

Titrations are carried out using a specialized piece of glassware called a buret, which is long tube with a dispensing valve. The buret scale has graduated marks in units of 0.01 mL or 0.02 mL. You can apply the techniques used for reading the graduated cylinder to reading a buret. The primary difference between a graduated cylinder and a buret is that the zero mark is at the top of the buret and at the bottom for a graduated cylinder.

The titrant solution is delivered through the stopcock (the dispensing valve) and bottom tip of the buret into a receiving flask. Therefore, the volume in the buret will decrease and the meniscus will drop as the solution is delivered. It is a common mistake to read the buret volume from the bottom-up. Simply read and record the number from the top down (to the right number of significant figures, and with units);. You don’t need to add or subtract your reading from 50.

In this experiment you will place the NaOH solution into a clean buret. The solution that is placed into the buret is known as the titrant. When readying the buret, over-fill the buret above the 0.00 mL mark. The stopcock is then opened to allow the solution

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Always measure the bottom of the meniscus at eye level.

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to flow out of the tip until the solution level falls just below 0.00 mL. This process ensures that the tip of the buret is filled with the solution and not with an air bubble (more on this later). It is not critical that the initial volume of the solution be exactly 0.00 mL. However, it is important that the value be between 0.00 and 1.00 mL, and that the exact value be recorded.

Once the buret is loaded with the titrant, we can then prepare the analyte solution in the Erlenmeyer flask. It is critical that an indicator be added to the analyte solution to signal the end point; otherwise, you will not know when to stop the titration and you may “over shoot” the end point.

Before the titration is started, the initial volume of titrant in the buret is read to the nearest 0.01 mL, i.e., to two decimal places; you estimate one decimal beyond the graduations. The titrant is added slowly to the analyte solution. As the addition continues, a fleeting color change may be noted (as a result of the indicator), especially when the titrant first contacts the analyte solution. It is important that the analyte solution be swirled constantly to ensure thorough mixing. Swirl carefully to avoid splashing of the analyte solution. The endpoint is not reached until the entire solution has a permanent color change (does not disappear with swirling). With practice, it is possible to determine the endpoint within a single drop of titrant solution. Once the color change is permanent, the first titration trial is complete, and the final volume can be read from the buret. The difference between the final and initial buret reading is the volume of titrant added to the flask.

The volume and concentration of the standard are used to calculate the moles of titrant added. The moles of titrant can then be stoichiometrically compared to the moles of analyte present, allowing us to calculate the concentration of the analyte solution.

Since titration is an analytical technique, multiple trials are run to ensure consistency of the results. Titrations should be run with a minimum of two trials; however, the more trials the better your results are likely to be. If, after performing a titration, you find that one set of results does not agree with the others, it should be excluded. Ideally, an additional trial should be run to confirm the results. Titration is an accurate and precise technique; if done correctly, it is possible to perform experiments with less than 1% error. It will take some practice for you to reach this level; nevertheless, you should strive to keep your errors low. Part of your grade will be based on the accuracy of your experimental results.

Consider the following example. A student performed an acid-base titration to determine the concentration of a potassium hydroxide solution, KOH(aq). The standard solution was 0.275 M nitric acid, HNO3. The student filled the buret above the 0.00 mark with the nitric acid solution, then drained it to 0.18 mL, ensuring the tip had no air bubbles. The student used a volumetric pipet to dispense exactly 25.00 mL of the unknown potassium hydroxide solution into an Erlenmeyer flask, and added a few drops of the acid base indicator. The student slowly added the nitric acid solution while swirling the Erlenmeyer flask, until the color change persisted throughout the solution, and recorded the final volume on the buret as 43.08 mL.

Volume of nitric acid used = Vfinal – Vinitial = 43.08 mL – 0.18 mL = 42.90 mL HNO3 To calculate the moles of nitric acid, use the relationship: moles = Molarity x volume in Liters

3 2-

3 3

3 HNO mol 10 x 1.18 mL 1000

L 1 x HNO mL 42.90 x HNO L 1

HNO mol 0.275 =

From the balanced chemical equation for the reaction between potassium hydroxide and nitric acid (equation 1), we know the stoichiometric relationship between potassium hydroxide and nitric acid is 1-to-1.

HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l) (1)

From the stoichiometry we can determine the moles of potassium hydroxide present.

KOH mol 10 x 1.18 HNO mol 1 KOH mol 1 x HNO mol 10 x 1.18 2-

3 3

2- =

The moles of potassium hydroxide and the volume of the solution used (25.00 mL = 2.500 x 10-2 L) can then be used to calculate the concentration of the potassium hydroxide solution.

M 0.472 mol/L 0.472 L 10 x 2.500

mol 10 x 1.18 KOH of L KOH mol solution KOH ofMolarity 2-

-2 ====

Each step (above) is shown individually for clarity. However, it is possible to combine all steps together using a factor- label sequence:

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KOH of L KOH mol solution KOH ofMolarity ==

mol/L 0.472 KOH L 1

KOH mL 1000 x KOH mL 00 25.

1 x HNO mL 1000

HNO L 1 x HNO mL 42.90 x HNO L 1

HNO mol 0.275 x HNO mol 1 KOH mol 1

3

3 3

3

3

3 =

Note that the key to a titration is the stoichiometric comparison of the moles of one reactant (the acid) to the moles of the other reactant (the base); without this comparison a titration is impossible.

A simpler solution to the above titration problem consists of using the molarity equation. The relationship is as follows.

B

BB

A

AA n

VM n

VM = (2)

The quantities are as follows: MA = molarity of the Acid; VA = volume of the Acid

MB = molarity of the Base; VB = volume of the Base

nA = coefficient for the Acid in the balanced chemical equation

nB = coefficient for the Base in the balanced chemical equation

In the acid-base reaction: HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l), nA = 1 and nB = 1; since the mole ratio of acid:base is 1:1.

Using equation 2, we can compute the molarity of the KOH, MB:

1

mL) (25.00 x M 1

mL) (42.90 x M) 275.0( B=

MB = molarity of potassium hydroxide = 0.472 M

Note that it is not necessary to convert the volumes to liters. Leave both the volumes in mL; these units will cancel.

PRIMARY STANDARDS As mentioned above, the standard solution must have a known concentration. Some compounds can be used to prepare solutions in which the concentration is accurately known. These compounds are called primary standards. It is important that a primary standard has the following characteristics:

• Purity – We need to be able to accurately determine the number of moles of primary standard present. Impurities of any kind make this difficult, if not impossible.

• Stability – Some compounds undergo chemical reactions while they are in the bottle or when exposed to air. Once this happens, the sample is no longer pure.

• High molar mass – Compounds that have a high molar mass are easier to weigh, thus reducing errors introduced in mass measurement. A small error in mass results in an even smaller error in the number of moles if the molar mass is large.

Fortunately, there are a number of primary standards that are not expensive or cumbersome to use. A small quantity of a primary standard is often used to standardize a large volume of solution. Once this solution is standardized, we call it a secondary standard, or simply, standard. This standard can be used in a titration (since its concentration is known) to determine the concentration of other solutions. When you use a secondary standard, you are performing two sets of titrations. The first set of titrations is used to standardize the secondary standard solution; the second set is to analyze the unknown sample.

In this experiment, you will use the primary standard potassium hydrogen phthalate (KHC8H4O4, abbreviated KHP) to standardize a sodium hydroxide solution. The stoichiometry for the reaction is 1-to-1.

KHC8H4O4(aq) + NaOH(aq) → KNaC8H4O4(aq) + H2O(l) (3)

HC8H4O4–(aq) + OH–(aq) → C8H4O42–(aq) + H2O(l) (4) Equation 4 is the net ionic equation. As you can see, one mole of KHP reacts with one mole of sodium hydroxide, and the potassium and sodium ions are spectator ions. Note that only one of the hydrogen atoms in KHP (shown in bold) is acidic and reacts with hydroxide ions. KHP is a monoprotic acid, meaning the compound contains only one acidic hydrogen atom.

The reason we need to standardize the sodium hydroxide solution is that NaOH pellets are hygroscopic, which means they readily absorb water from the air, resulting in an impure sample as soon as the bottle is opened. Therefore, when

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you weigh a sample of sodium hydroxide, you are actually determining the mass of a mixture, NaOH and H2O. It is very important that students immediately replace the cap on the sodium hydroxide bottle to minimize the absorption of water. Sodium hydroxide exposed to air will eventually absorb enough water to form a solution. To determine the concentration of the NaOH solution, we need to know the moles of KHP used and the volume of NaOH used to neutralize all the KHP. If we know the mass and the molar mass of KHP, we can calculate the moles of KHP. The stoichiometry is 1-to-1 so this is also equal to moles of NaOH. Molarity (concentration) is moles of solute (NaOH) per liter of solution (the volume of NaOH used). Suppose 1.2683 g of KHP (MM = 204.22 g/mol) was titrated with 14.67 mL of NaOH(aq). The molarity of the NaOH(aq) solution can be determined as follows:

moles of KHP = KHP g 204.22

KHP mol 1 x KHP g 1.2683 = 6.2105 x 10–3 mol KHP

moles KHP = moles NaOH = 6.2105 x 10–3 mol NaOH

M 0.4233 mol/L 0.4233 L 10 x 1.467 mol 10 x 6.2105

NaOH of L NaOH mol solution NaOH ofMolarity 2-

-3 ====

TITRATION OF VINEGAR Once we have standardized our NaOH(aq) solution, we can use it to determine the concentration of an unknown acid. In this experiment you will titrate a sample of vinegar (the analyte) with the NaOH (the titrant). Vinegar is a dilute solution of acetic acid, CH3COOH; a weak monoprotic acid and therefore a weak electrolyte containing a small concentration of ions. The reaction with NaOH is:

CH3COOH(aq) + NaOH(aq) → CH3COONa + H2O(l) (5)

CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l) (6) You will already have determined the concentration of the NaOH(aq) solution. You will know the volume of vinegar used and the volume of NaOH(aq). Using the molarity equation (equation 2), it is then quite straightforward to compute the molarity of the vinegar solution.

MASS PERCENT CALCULATIONS Some commercially available solutions, such as vinegar, have a concentration unit of mass percent (m/m %) on the label.

For a solution, mass percent = 100x solution of mass solute of mass

We can convert between concentration units of mass percent and molarity (M). For example, a solution that is labeled as 37% HCl, means there are 37 g of HCl for every 100 g of solution. We need to know the density of the solution, which in this case is 1.19 g/mL, and the molar mass of HCl, which is 36.46 g/mol.

HCl g 36.46 HCl mol 1 x

solution L 1 solution mL 1000 x

solution mL 1 solution g 1.19 x

solution g 100 HCl g 37

solution of L HCl mol solution HCl ofMolarity ==

= 12 mol/L = 12 M

For the reverse calculation (molarity to mass percent), the density, molar mass, and molarity are still needed for this calculation:

100 x solution g

HCl g solution HCl of percent Mass =

100 x solution g 1.19

solution mL 1 x solution mL 1000

solution L 1 x HCl mol 1 HCl g 36.46 x

solution L 1 HCl mol 12 =

= 37 %

In this experiment you will use a density of the vinegar solution of 1.02 g/mL, along with your experimentally-determined value of the molarity, to calculate the mass percent, and compare your value to that on the bottle.

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AIR BUBBLES An air bubble in the tip of your buret, or anywhere in the buret, can be a serious problem. Imagine that an air bubble in the buret occupies a volume of 1.00 mL (this is a large bubble, but the exaggerated volume helps illustrate the problem). Assume that a student makes an initial volume reading at 0.22 mL and reaches an end point at 28.72 mL. The student’s calculated volume is 28.50 mL.

DV = Vfinal – Vinitial = 28.72 mL – 0.22 mL = 28.50 mL

However, 1.00 mL of the volume dispensed was due to an air bubble and not the titrant solution. Therefore, only 27.50 mL of the titrant was added to the analyte. The volume has an error of >3%. If the air bubble had been removed before the titration, the error could have easily been avoided. Since titration requires precision, assuming it is carried out correctly, every precaution should be taken to reduce errors. With practice, it is possible to get results with <1% error.

CHECKING YOUR WORK If you are performing the standardization of base accurately (part A), the molarities that you calculate from multiple trials (at least two) should agree. Similarly, the molarities of vinegar that you measure in multiple trials in part B of the experiment should agree closely. It is your responsibility to make certain that you have at least two good trials (good means you are confident you made no errors, and the results of the two trials agree to within 1%) for each part of the experiment.