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pH OF SOLUTIONS
OBJECTIVES
- To investigate the strengths of acids and bases
- To examine the effect of concentration on the pH of a solution
- To examine the effect of salt hydrolysis on pH
- To determine the properties of buffer solutions
INTRODUCTION
According to Bronsted/Lowry theory, an acid can be defined as a substance that donates a proton, or hydrogen ion, to another substance. A base is a substance that accepts a proton. In aqueous solution, protons will bond with water molecules to form hydronium ions (H 3 O+). Thus in an aqueous solution of HCl, HCl is the acid and water is the base. Similarly, in an aqueous solution of acetic acid, CH 3 COOH is the acid and water is the base. But in an ammonia solution, water is the acid and ammonia is the base. Review conjugate acid/base pairs in your book and look carefully at the equations below to be sure you understand this relationship.
HCl(aq) + H 2 O(l) 6 H 3 O+(aq) + Cl&(aq) (1)
acid 1 base 2 acid 2 base 1
CH 3 COOH(aq) + H 2 O(l) ∫ H 3 O+(aq) + CH 3 COO&(aq) (2)
acid 1 base 2 acid 2 base 1
NH 3 (aq) + H 2 O(l) ∫ NH 4 +(aq) + OH&(aq) (3)
base 1 acid 2 acid 1 base 2
As you can see, water can act as an acid or a base. It can also undergo a process called auto- ionization , represented by the equation below.
H 2 O(l) + H 2 O(l) ∫ H 3 O+(aq) + OH&(aq) (4)
This reaction is an equilibrium, with the equilibrium lying far to the left. Since the concentration of water remains essentially constant, the equilibrium expression can be written as
Kw = [H 3 O+][OH&] (5)
where Kw equals 1.0 x 10-14^ at 25oC. In pure water or in any neutral aqueous solution at 25oC, the concentration of H 3 O+^ ions must be equal to the concentration of OH&^ ions. Both are 1.0 x 10-7M. In an acidic solution , [H 3 O+] > [OH&]. In a basic or alkaline solution , [OH&] > [H 3 O+]. However, the product must always be 1.0 x 10-14^ at a temperature of 25oC.
Scientists often report the acidity of a solution in terms of pH , where pH is defined as the negative logarithm of the hydronium ion concentration, or
pH = & log 10 [H 3 O+] (6)
For any neutral solution, pH = & log (1.0 x 10-7M) = 7.0. Thus in an acidic solution, the pH must be less than 7, while a basic solution must have a pH greater than 7. A lower pH indicates a greater concentration of hydronium ions and a more acidic solution; a higher pH indicates a lower concentration of hydronium, a greater concentration of hydroxide, and thus a more basic solution.
Part I. Strength of acids and bases
Acids are defined as strong or weak depending on their extent of ionization in aqueous solution. Strong acids ionize essentially completely in dilute solution to form H 3 O+^ ions; weak acids ionize only slightly. Look again at equations (1) and (2). Notice that the arrow is shown going in only one direction for the HCl reaction, indicating extensive ionization; however, the reaction between acetic acid and water is reversible. Thus HCl ionizes essentially completely while acetic acid ionizes to a much smaller extent. This difference in ionization will be reflected in a difference in pH. Base strength is defined in a similar manner. Strong bases such as NaOH extensively dissociate in solution, forming hydroxide ions.
NaOH (aq) 6 Na+(aq) + OH&(aq) (7)
Weaker bases, such as ammonia, ionize only slightly and produce a lower concentration of hydroxide ions, as shown in equation (3).
Part II. Concentration and pH
If the concentration of an acid or base in solution is changed, the concentration of hydronium ions and hydroxide ions will also change. For example, if the concentration of hydronium ions increases, hydroxide ion concentration must decrease to maintain the equilibrium shown in equation (4). This will be reflected in a change in the pH of the solution.
Part III. Salt hydrolysis
Salts are formed from the reaction of an acid with a base. A salt consists of a cation (which is the conjugate acid of the base that reacted) and an anion (which is the conjugate base of the acid that reacted). Solutions of some salts are neutral, but other salts will react with water in what is known as a hydrolysis reaction. When this happens, the salt solution will not have a pH of 7.
We can identify different categories of salts, based on the strength of the corresponding acids and bases:
- A salt formed from reaction of a strong base and a strong acid should give a neutral solution. For example, KBr is formed from the reaction of KOH, a strong base, and HBr, a strong acid. Neither the potassium ion nor the bromide ion can undergo a hydrolysis reaction, so a potassium bromide solution should be neutral. (NOTE: Often solutions of pure water or neutral salts are slightly acidic, due to the formation of carbonic acid from the reaction of carbon dioxide from the air with water.)
CO 2 (g) + H 2 O(l) W H 2 CO 3 (aq) W H+(aq) + HCO 3 G(aq) (8)
PROCEDURE
You should obtain 6 test tubes from the instructor. Clean them thoroughly and rinse well with distilled water as needed for the remaining parts of this lab. Prepare each tube and then measure its pH using pH paper. Record pH on the laboratory report sheet. NOTE: Before you discard the contents of any tube, check to see whether it is needed for the next section.
Part I. Strength of acids and bases
Tube 1: 2 mL of 0.10M HCl (Volume can be approximated for this section.) Tube 2: 2 mL of 0.10M acetic acid Tube 3: 2 mL of 0.10M NaOH Tube 4: 2 mL of 0.10M NH 3
Part II. Concentration and pH
Tube 5: 1 mL of 0.10M HCl from tube 1 thoroughly mixed with 9 mL distilled water Tube 6: 1 mL of the solution from tube 5 thoroughly mixed with 9 mL distilled water Tube 7: 1 mL of 0.10M NaOH from tube 3 thoroughly mixed with 9 mL distilled water Tube 8: 1 mL of the solution from tube 7 thoroughly mixed with 9 mL distilled water
Part III. Salt hydrolysis
Tube 9: 2 mL of 0.10M NaCl (Volume can be approximated for this section.) Tube 10: 2 mL of 0.10M NaCH 3 COO Tube 11: 2 mL of 0.10M NaNO 3 Tube 12: 2 mL of 0.10 M NH 4 Cl
Part IV. Buffers
Tube 13: 2 mL of 0.20M CH 3 COOH thoroughly mixed with 2 mL of 0.20M NaCH 3 COO
After you have measured the pH of tube 13, divide its contents equally between 2 tubes. These will now be tubes 14 and 15.
Tube 16: 2 mL of 0.20M NH 3 thoroughly mixed with 2 mL of 0.20M NH 4 Cl
After you have measured the pH of tube 16, divide its contents equally between 2 tubes. These will now be tubes 17 and 18.
Tube 19: 2 mL distilled water Tube 20: 2 mL distilled water
Arrange tubes 14, 17, and 19 in a rack. To each, add 2 drops of 0.10M HCl. Mix thoroughly and record the pH. Add 2 more drops of the HCl; mix and record the pH. Continue the addition as described until you have added a total of ten drops.
Arrange tubes 15, 18, and 20 in a rack. To each, add 2 drops of 0.10 M NaOH. Mix thoroughly and record the pH. Add 2 more drops of the NaOH; mix and record the pH. Continue the addition as described until you have added a total of ten drops.
Part IV. Buffers
Calculate the concentrations in each tube after mixing and record along with the measured pH. Show your work in the space provided below.
TUBE # CONTENTS CONCENTRATIONS MEASURED pH
13 CH 3 COOH NaCH 3 COO
16 NH 3 NH 4 Cl
TUBE # CONTENTS MEASURED pH AFTER ADDING HCl 2 drops 4 drops 6 drops 8 drops 10 drops
14 CH 3 COOH NaCH 3 COO
17 NH 3 NH 4 Cl
19 Water
TUBE # CONTENTS MEASURED pH AFTER ADDING NaOH 2 drops 4 drops 6 drops 8 drops 10 drops
15 CH 3 COOH NaCH 3 COO
18 NH 3 NH 4 Cl
20 Water
QUESTIONS
Part I.
- Why is the pH of 0.10 M acetic acid not the same as the pH of 0.10 M HCl?
- Which solution has the higher pH, 0.10 M NaOH or 0.10 M NH 3? Is this the result you expected? Why or why not?
Part II.
- Compare your results for tubes 1, 5, and 6. What is the relationship between the concentration of HCl and the pH?
- Compare your results for tubes 3, 7, and 8. What is the relationship between the concentration of NaOH and pH? Is this what you expected? Explain.
