# Solutions and pH

###### LEARNING OBJECTIVES

The students will be able to:

1. Learn how to prepare aqueous solutions.

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2. Learn to dilute solutions.

3. Measure the pH of a solution.

###### BACKGROUND

Solutions are homogeneous mixtures. Solutions have two parts: the solute (the dissolved substance) and the solvent (the one that dissolves the solute). In aqueous solutions, the solvent is distilled or deionized water. Tap water cannot be used because it contains dissolved solutes. Pure water does not conduct electricity; it is a non-electrolyte. Water has high dielectric constant to break the electrostatic interactions between ions with opposite charges. In solution, the ions can travel freely and they can conduct electricity. The types of solutes, where all the dissolved molecules are fully dissociated, are called strong electrolytes. Solutions can usually be prepared by dissolving a solid substance in water. The quantitative description of a solution is expressed by molarity. The molarity of a solution is the number of moles of solute in a liter of solution.

= = molarity; = number of moles; = volume in liter

Example 1: A solution contains 11.70 g of Sodium Chloride in 250.0 mL of solution. What is the molarity of the solution? Na = 23; Cl = 35.5

Solution:

NaCl = 58.5 g/mol and V = 0.2500 L 0.200 mol/L.

= 0.800 mol/L.

Example 2: A sample of 4.90 g of CuSO4.5H2O is dissolved in enough water to prepare 400.0 mL solution. What is the molarity of the solution? Cu = 63.54; S = 32.00; O = 16.00; H = 1.01 Solution: The formula weight of is CuSO4.5H2O = 249.5 g/mol.

= 0.0196 mol. 0.0491 mol/L.

Sometimes, a solution can be highly concentrated (high solute content) and needs to be diluted. The amount of solute remains the same, however by adding more solvent the concentration of the solution will be lowered. The formula used to calculate the new concentration is:

Mi × Vi = Mf × Vf

Note: i for initial and for final.

Some substances, such as HCl and HNO3, when dissolved in water release H+ ions into solutions. These solutions are acidic and have a pH < 7.0. Others, such as NaOH and KOH, release OH- ions into solutions. These solutions are basic and have a pH > 7.0. Solutions that have a pH = 7.0 are called neutral. A neutralization reaction is a reaction between an acid and a base to produce a salt and water.

In summary, the pH of a solution can vary as follows:

 Acidic: 0 pH < 7.0 Neutral: 7.0 Basic: 7.0 < pH 14.0

Example 3: What volume of water is needed to dilute 50.0 mL 12 HCl solution in order to obtain 3 HCl solution?

Solution: (12 M) × (0.050 L) = (3 M) × (Vf) 0.60 = (3 M) × (Vf)

Vf = 0.200 L = 200 mL.

The volume of water needed is 200 – 50 = 150 mL

There are two parts in this laboratory experiment. In the first part, you will prepare 100.0 mL of a 0.10 M solution of copper(II) sulfate using solid CuSO4.5H2O. Then, you will prepare more solutions by

diluting this stock solution. In the second part, you will prepare solutions of acids and bases and determine their pH’s.

###### SAFETY PRECAUTIONS

1. No drinking or eating in the laboratory. You have follow all laboratory safety guidelines.

2. Obtain and wear chemical-splash eye goggles at all times.

3. Dilute acids and bases are irritating to eyes and skin.

4. After the experiment is finished, clean up and wash hands thoroughly with soap and water beforeleaving the laboratory.

###### MATERIALS

NaCl solution, 0.10 M

CuSO4.5H2O, solid HCl and NaOH, 0.10 M

Phenolphthalein Deionized water Balance

Well-plates

Toothpicks

Small test tubes

(5) Plastic pipets pH paper Beakers 50-mL

Graduated cylinders, 10 and 25- mL

PRELAB QUESTIONS NAME:

1. Define the units for molarity.

2. A solution contains 5.25 grams of sodium chloride in 750.0 mL of solution. Determine the molarity of the solution. Na = 23.00, Cl = 35.54

3. A 7.59 gram of sample of CuSO4.5H2O is dissolved in enough water to prepare 300.0 mL solution.

Determine the molarity of the solution. Cu = 63.54, S = 32.00, O = 16.00, H = 1.01

4. What volume of water is needed to dilute of a 10.0 mL 5.0 sugar solution to obtain a 0.25 M

sugar solution?

DATE: NAME:

DATA SHEET FOR PROCEDURE A.

Part A. Preparation of stock solution and making serial dilutions

###### PROCEDURE
 1. Weigh a 50-mL beaker. Record the mass. g

138

 2. Add about 0.5 gram of CuSO4.5H2O. Record the mass.     3. Mass of CuSO4.5H2O g (Line 2- Line 1)     4. Determine the number of mol of CuSO4.5H2O. g ((Line 3) / 249.5 g/mol) mol

5. Add some deionized water to the beaker. Carefully mix with a stirring rod and transfer its contents to a 100mL volumetric flask #1.

6. Repeat step 5 until all solid is dissolved. Then fill the volumetric flask to the mark with deionizedwater (100.0mL = 0.100 L).

7. Calculate the molar concentration. (line 4 / 0.100)

mol/L

8. Pipette 10.00 mL of solution from volumetric flask #1 and transfer it to volumetric flask #2 Fill to themark with deionized water. Determine the molarity of the new solution.

M.

9. Pipette 10.00 mL of solution from volumetric flask #2 and transfer it to volumetric flask #3. Fill volumetric flask #3 to the mark with deionized water. Determine the molarity of the new solution.

M

10. Pipette 10mL of solution from volumetric flask #3 and transfer it to volumetric flask #4. Fill volumetric flask #4 to the mark with deionized water. Determine the molarity of the new solution.

M

11. Obtain four small test tubes. Arrange them in a row. Fill each test tube, in turn, with the solution from volumetric flask 1, 2, 3, and 4.

12. Take a picture of the test tubes with your smartphone. (BE CAREFUL TO NOT WET YOUR PHONE!)

12. Obtain the CuSO4 solution with an unknown concentration. Compare the color visually or to the picture taken with your phone and evaluate its concentration.

The concentration of the unknown solution is M

DATE: NAME:

#### PART B.

###### THE pH’s OF ACIDS AND BASES PROCEDURE A.

1. Obtain three small test tubes. Add 1.0 mL of 0.10 HCl to the first test tube. Add 1.0 mL of 0.10 NaOH to the second test tube and 1.0 mL of 0.10 NaCl solution into the third test tube.

2. Using a pH stick, measure the pH of each solution and record them. A. pH of 0.10 M

HCl solution .

B. pH of 0.10 NaOH solution . C.

pH of 0.10 NaCl solution .

3. Add one drop of phenolphthalein indicator to each of the test tube and record the colors.

A. Color of 0.10 HCl solution .

B. Color of 0.10 NaOH solution .

C. Color of 0.10 NaCl solution .

###### PROCEDURE B.

1. Obtain a small test tube and using a plastic pipet add 5 drops of 0.10 of HCl and 1 drop of phenolphthalein indicator solution. Add 10 drops of 0.10 of HCl and 1 drop of phenolphthalein to a second small test tube. Add 15 drops of 0.10 of HCl and 1 drop of phenolphthalein to a third small test tube. The total drops of acid is 30 drops in three test tubes.

2. Using a disposable pipette, to each tube, add 0.10 NaOH solution until it turns pink. Record the number of drops of NaOH for each:

Test tube 1. 5 drops of HCl drops of NaOH

Test tube 2. 10 drops of HCl drops of NaOH Test tube 3. 15 drops of HCl drops of NaOH

Total 30 drops of HCl drops of NaOH

3. Compare the total number of drops of NaOH to the total number of drops (30 drops) of HCl.

4. HCl is titrated with NaOH. Actual values (number of drops) should be equal. Explain why?

POSTLAB QUESTIONS NAME:

1. In the experiment PART B did you achieve neutralization with 30 drops of 0.10 NaOH solution? If not, explain why.

2. What mass of K2SO4 is needed to prepare 500.0 mL of a 0.75 M K2SO4 solution? K = 39; S

= 32; O =16.

3. What volume of water is needed to dilute a 25 mL of a 5.0 sugar solution to obtain a 1.0 M

sugar solution?

4. 50.00 mL of a 0.25 NaCl solution is added to 150.00 mL of a 0.50 NaCl solution. What is the molarity of NaCl in the final solution?

5. A 10.00 mL of a 0.20 NaOH solution is added to 10.00 mL of a 0.10 HCl solution containing one drop of phenolphthalein indicator. What color change is observed?

6. A 10.00 mL sample of an HCl solution requires 40.00 mL of a 0.25 M NaOH solution to achieve neutralization. What is the molar concentration of the HCl solution?

NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)