pH Measurement-Relative Molar Mass, Chemistry tutorial

Introduction

For most of the science experiments, we will require a pH indicator, these as wide-range litmus, pH paper, or pH meter. Such pH indicators have a chemical that transforms colour when it comes in contact through acids or bases. For instance, litmus and pH paper turn red in strong acids and blue in strong bases. Since only a few pH indicators determine pH over a broad range of pH values, we will require finding out the pH range of the indicator we utilize. Classically, the colour chart presented through each pH indicator kit will demonstrate the pH range of that indicator. Colour pH indicators give only an estimated determine of the pH, or the strength of the acid or base. They aren't as precise as the expensive instruments scientists utilize to calculate pH, but they are adequate for the experiments we require to do at this level. 

Materials

pH paper and colour chart (pH range 3 to 12) or pH meter

Distilled water, white vinegar, household ammonia (or baking soda)

Spot plate test or 3 small test tubes, stirring rod

Solutions / fruits juice (lemon, lime, orange, or melon), beverages (cola, carbonated non-cola, milk)   

Indicators, in chemistry, are natural or synthetic substances that modify colour in answer to the nature of chemical environment. Litmus, for instance, is a natural dye that turns red in most acidic solutions and blue in most basic solutions. Compounds that undergo colour transforms when there is a pH change in the solutions in that they are enclosed are termed indicators (Table).

Indicators are utilized to provide information about the degree of acidity of a substance or the state of several chemical reactions within a solution being tested or examined.

How to Measure with ph Paper

When measuring pH with pH paper, dip the end of a strip of pH paper into each mixture you want to test. After about 2 seconds,  eliminate the paper, and instantly compare the colour  at  the  wet end of  the  paper  by  the colour chart provided  through  that pH indicator. When measuring pH with pH meter, dip the end of electrode of pH meter into each mixture we want to test. Write down the pH value and colour. Always utilize a clean, unutilized strip of pH paper for each mixture that we test.

Indicators

Acid(colour change)

Base(colour change)

pH range

Methyl orange

Red

Yellow

3.1-4.4

Methyl Red

Red

Yellow

4.2-6.3

Bromothymol Blue  

Yellow

Blue

6.0-7.8

Phenol Red

Yellow

Red    

6.4-8.0

Phenolphthalein

Colourless

Pink

8.0-9.8

Thymol Blue

 Red

Yellow

1.2-2.8

Alizarin Yellow

Yellow

Red

10.1-12.0

Table: List of indicators and their pH range

Example 1 - Vinegar is an acid, and in this experiment it will show a pH of about 4.

Vinegar at pH 4 rotates pH paper yellow and most other pH indicators red.

Example 2 - Ammonia is a base and in this experiment it will display a pH of about 12. Bases turn most pH indicators blue.

Example 3 - Pure distilled water would have tested neutral, but pure distilled water is not easily obtained because carbon dioxide in the air around us mixes, or dissolves, in the water, making it somewhat acidic. The pH of distilled water is between 5.6 and 7. Traditionally, solutions were labeled as being acidic or basic based on their taste and texture. Those that tasted sour were said to be acidic and solutions that tasted bitter and were slippery to touch were said to be basic. Therefore, substances such as lemon juice and vinegar were identified as acids, and solutions of lye and caustic soda as bases. Numerous definitions have been proposed for acids and bases. Depending upon the situation, one or more definition is applicable.

In this case, the hydronium ion, H3O+, shapes when a proton, H+, is transferred from one H2O molecule to another. The other species that result from this process is the hydroxide ion, OH-. Therefore while one water molecule, the proton acceptor, functions as the base; the other plays the role of the acid. The ability of one water molecule to accept a proton from another water molecule or any acid is due to the 2 lone pairs of electrons on the oxygen atom of water.

The auto ionization consequences in equivalent molar amounts of H3O+ ions and OH- ions and hence the solution is neutral. In an instance of pure H2O, the concentrations of H3O+ and OH- ions at 25oC are 1.0 x 10-7 M.

[H3O+] = [OH‾] = 1.0 x 10-7 M

The concentration of hydronium ion, [H3O+], of a solution is commonly expressed in expressions of the pH of  the solution that is described as the negative  logarithm of [H3O+] or negative logarithm of [H+] in the solution;

pH = - log [H3O+] or pH = - log [H+]

Therefore  the  hydrogen ion concentration  can be  attained from  the  pH  of the  solution  as follows;

[H3O+]  = 10-pH

Similarly the concentration of hydroxide ion, [OH-], of a solution is generally expressed in terms of the pOH of the solution that is described as the negative logarithm of [OH-]

pOH = - log [OH ‾]

The hydroxide ion concentration can be attained from the pOH of the solution using the equation

[OH‾] = 10 ‾ pOH

Additionally, the pH and the pOH of any aqueous solution are related as are the hydrogen and the hydroxide ion concentrations. The relevant equations are;

[H+] [OH‾] = 1.0 x 10-14 pH + pOH = 14

Example (1)

What is the pH and pOH of a solution that contains 3.50 x 10-5 M hydronium ions?

pH = - log [H3O+]  = - log (3.50 x 10-5)  = 4.46 pOH = 14 - pH = 14 - 4.46 = 9.54

Example (2)

Calculate the hydronium ion and hydroxide ion concentrations of a solution that has a pOH of 4.40. pH = 14 - pOH = 14 - 4.40 = 9.60 [H3O+] = 10 - pH  = 10 -  9.60   = 2.51 x 10-10 M

[OH-] = 10 - pOH = 10 - 4.40   = 3.98 x 10 -5 M

In water, [H3O+] is equal to 1.0 x 10-7

M, so the pH is 7.0. Because [H3O+] = [OH-] in water, which is neither acid nor base. pH = 7.0 (neutral)  pH < 7.0 (acidic)  pH >  7.0 (basic)

The pH scale has a range of 0.0 to 14.0. A practical way to evaluate the relative acidity or basicity of solutions is to compare their effect on indicators. In this experiment, we will examine the properties of acids and bases through appropriate indicators. By the end of this experiment, we will be able to determine the pH of various solutions such as various fruits, common beverages, and borax. Clorox and Borax are cleaning agent that several people add to their laundry detergent. Acids generally taste sour, and bases bitter. Household cleaners are poisons so we should never taste them. We will examine the colour of several indicators in such solutions and as well using pH paper and pH meter.

Procedure

1. Dip an unused strip of pH paper into solution. Leave until wet (about 2 seconds). Instantly compare through the color chart.  Write down the estimated pH value of the solution. If we're using pH meter write down the approximate pH value of the solution.

2.  Repeat the similar procedure for the other solutions and record our observation in a table.

Description on relative molar mass:

The vapour pressure of a pure liquid at a following temperature is a trait property of that liquid. Though, whenever a nonvolatile solute is dissolved in the liquid, the vapour pressure of the liquid is decreased. This lowering of the vapour pressure reasons a transform in the melting point, boiling point, and osmotic pressure of the liquid. The magnitude of the change in such properties based upon the number of solute particles dissolved in a following amount of the solvent, but not upon the nature of the particles (their identity). These properties are called colligative properties. The addition of ethylene glycol to the water in a car radiator in order to increase its boiling point, or utilize of salt to lower the melting point of ice on a sidewalk is various everyday applications of colligative properties.

Theory

Most substances can exist as solid, liquid, or gas, depending upon the temperature and pressure. Whether a particular substance exists as a solid, liquid, or gas under states of standard temperature and pressure is dependent upon the nature of the substance and contains these properties as the molecular weight of the substance or the intermolecular forces of attraction between molecules.

If a substance that exists as a solid is heated, it will ultimately melt to form a liquid. Expect a substance, which is a solid at room temperature is gradually heated. As energy is added to the solid, the temperature of the solid will begin to increase. Once the melting point of the solid is reached, though, the temperature of the solid-liquid mixture will continue steady until all the solid has been transferred into a liquid. Only then will the temperature increase again.  Does this make sense? If we are continuing to add heat to the example, how can the temperature remain constant? The answer to this question is that the energy being added is utilized to bring about the phase modify rather than to heat the example. The procedure of melting engages a breakdown of the attractive forces between molecules and needs energy.

A similar procedure take place as a liquid example is cooled. As heat is eliminated from the liquid, the temperature of the liquid will drop until the freezing point of the liquid is attained. Once the liquid starts to solidify, the temperature of the liquid-solid mixture continues steady until all of the liquid has solidified, and only then will the temperature start to drop again.  This is the cause why oranges are sprayed through water if a freeze if supposed; the temperature of a water-ice mixture can't drop below the freezing point of water until all of the water has frozen.  It is important to note that freezing and melting is really the similar thing and occurs at the same temperature for a particular substance. The freezing point of a liquid is depressed when it contains a dissolved solid. A solution of salt water, for instance, will freeze lower than the normal freezing point of water. The freezing point depression, or the difference between the freezing points of the pure solvent and solution, depends upon the number of particles in solution. The greater the concentration of the solution, the greater will be the freezing point depression. For a given concentration, a solute that dissociates will as well bring about a greater freezing point depression. For instance, a solution of NaCl will have twice the freezing point depression of a solution of sucrose of the similar concentration. The sucrose is a molecular substance that doesn't dissociate in solution, but the NaCl will dissociate into Na+ and Cl- ions in solution, giving twice as many particles in solution. The freezing point depression of a solution is symbolized through the subsequent equation:  ΔT = Kfm

Where:

ΔT = the freezing point depression

Kf = the freezing point depression constant

m= the molality of the solution, or the number of moles of solute per kilogram of solvent.

The freezing point depression constant, Kf, is different from solvent to solvent. Hence a given quantity of solute will not always bring about the similar freezing point depression. Several typical values are listed in the table below. We can see from this table that, of the substances listed, naphthalene has the largest value. What this means is that a 1.0 molal solution of a nondissociating solute dissolved in naphthalene would freeze 6.8 degrees below the normal freezing point of napthalene.

Solvent

Kf (oC/m)

Water

1.86

Benzene

5.12

Napthalene

6.8

Chloroform

4.68

Cyclohexane

20.4

The equation above can be exploited in numerous ways. Suppose a recognized quantity of solute is dissolved in an identified quantity of solvent and the freezing point depression is determined. If the molecular weight of the solute is known, the only unknown variable is the freezing point depression steady. If the freezing point depression steady is known, the equation can be solved for the number of moles of solute, from which the molar mass of the solute can be computed. This is the approach to be taken in lab today. In the 1st part of the experiment, the freezing point of pure paradichlorobenzene will be measured. In the 2nd part of the experiment, the freezing point depression constant for paradichlorobenzene will be determined through adding a known quantity of napthalene and measuring the freezing point of the resulting solution. In the last part of the experiment, the molar mass of an unknown solute will be determined based upon the freezing point of a solution.

Procedure

Determining the freezing point of pure paradichlorobenzene. Set up a hot water bath using a tripod, burner, and an 800 or 1000 mL beaker. Attain the mass of an empty large test tube. Fill the test tube approximately one-fourth to one-third full of solid paradichlorobenzene and again determine the mass of the tube. Subtract the 2 masses to attain the mass of paradichlorobenzene. Clamp the test tube in situate in the hot water bath. Once all the paradichlorobenzene has melted, eliminate the tube from the water bath and insert a rubber stopper enclosing a thermometer and wire stirrer. Stirring the instance constantly, record the temperature of the sample every 30 seconds. Graph of our consequences and record the melting point of pure paradichlorobenzene.

Determination of the freezing point depression steady. Weight out approximately 0.50 grams of naphthalene, C10H8, on a small piece of weighing paper. Return the test tube to the hot water bath and once the example has entirely melted, add the naphthalene. Eliminate the test tube from the hot water bath and repeat the above procedure. Compute the freezing point depression of the solution, the molality of the solution, and from this data the freezing point depression constant for paradichlorobenzene. Return the test tube to the hot water bath and heat until the instance has melted. Pour the melted solution into the 'PDB WASTE' container. If any solid remains in the test tube, wash it by a tiny portion of acetone. Any acetone waste should as well go into the waste container. We might wish to replicate this process a second time.

Determination of the molar mass of an unknown solute. Refill the test tube with fresh paradichlorobenzene and determine the mass of the test tube and contents. Weight out approximately 0.50 grams of unknown solute on a small piece of weighing paper. Add this to the test tube. Heat the test tube in the hot water bath until the mixture as completely melted. Remove the test tube from the water bath, insert the stopper containing the thermometer and wire stirrer, and record the temperature of the solution as in the preceding trials. Determine the freezing point of the solution, the freezing point depression, and from this data calculate the molar mass of the unknown solute. Reheat the test tube and pour the molten solution into the 'PDB WASTE 'container. If the tube doesn't come entirely clean, wash through a tiny portion of acetone. We might wish to replicate this process a second time.

For your reports:

1. Make 3 graphs-a cooling curve pure paradichlorobenzene versus time, a cooling curve of for the paradichlorobenzene-naphthalene mixture versus time, and a cooling curve of the paradichlorobenzene-unknown solute mixture versus time

2. Compute the freezing point depression steady for paradichlorobenzene

3. Compute the molar mass of the unknown solute

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