Lab 6 - Mixtures of Acids and Bases

Purpose

To investigate the resulting pH's of different mixtures of acid and base solutions.

Goals

• •
  To calculate the pH of pure acid and base solutions.
• •
  To calculate the pH of mixtures of acid and base solutions.
• •
  To measure the pH of pure acid and base solutions and mixtures of these solutions.
• •
  To compare the resulting pH's of different mixtures of acid and base solutions.

Introduction

Brønsted acids are defined as proton donors. When placed in water, Brønsted acids (HA) donate a proton to water to produce the conjugate base of the acid

(A⁻) 

and a hydronium ion (H₃O⁺) according to the following reaction. This is defined as the acid dissociation reaction and the equilibrium constant for this reaction is called the acid dissociation constant, K_a. K_a's can be found in the tables of physical data, and may be found in an appendix in your textbook. The larger the K_a for an acid, the stronger the acid is. The Brønsted acids are divided into two main groups: strong acids and weak acids.

Strong Acids:

The K_a's for strong acids, such as HCl and HNO₃, are simply listed as >> 1 which indicates that they dissociate 100% when placed in water. Since strong acids dissociate completely, the [H₃O⁺] = [strong acid concentration] and

pH = −log[H₃O⁺]. 

Weak Acids:

Weak acids, such as NH₄⁺ and HC₂H₃O₂, only partially dissociate in water according to their K_a's. As with any equilibrium reaction, the equilibrium concentrations can be calculated by generating a reaction table and substituting into the K_a expression. For example, acetic acid, HC₂H₃O₂, has a K_a of

1.8 × 10⁻⁵. 

So for a 0.2 M aqueous solution of HC₂H₃O₂, the equilibrium concentrations can be calculated as follows.

	HC2H3O2	+	H2O		C2H3O2−	+	H3O+
initial	0.2 M				0.0 M		0.0 M
Δ	−x				+x		+x
equilibrium	0.2 − x M				x M		x M

Thus the equilibrium concentrations are

[HC₂H₃O₂] = 0.2 M 

and

[C₂H₃O₂⁻] = [H₃O⁺] = 1.9 × 10⁻³ M. 

Equation 3

Ka = [C2H3O2− ][H3O+] [HC2H3O2] = (x)(x) (0.2 − x) ≅ x2 0.2 = 1.8 × 10−5

x = 1.9 × 10−3 M

 

shows the detailed method to calculate [H₃O⁺] for a weak acid. In cases where the percent ionization of the acid is less than 5%, equation 3

Ka = [C2H3O2− ][H3O+] [HC2H3O2] = (x)(x) (0.2 − x) ≅ x2 0.2 = 1.8 × 10−5

x = 1.9 × 10−3 M

 

simplifies to the expression below, where c_wa represents the original concentration of the weak acid in M. The pH of this solution can now be calculated by taking the

−log[H₃O⁺] 

to get 2.72. Similarly, Brønsted bases

(B⁻) 

are defined as proton acceptors and when placed in water will accept a proton from water to produce the base's conjugate acid (HB) and a hydroxide ion

(OH⁻). 

This reaction is defined as the base association reaction and the equilibrium constant for it is the base association constant, K_b. K_b's cannot be found in tables as K_a's are. K_b's must be calculated according to the following equation at 25°C. The Brønsted bases are also divided up into two main groups: strong bases and weak bases.

Strong Bases:

The most common strong bases found in the laboratory are the Group IA salts of hydroxide, which completely dissociate in water to give one equivalent of

OH⁻. 

Since the salts dissociate completely, the

[OH⁻] = [strong base concentration] 

and the

pOH = −log[OH⁻]. 

At 298 K, the relationship between pH and pOH is:

Weak Bases:

Sodium acetate, NaC₂H₃O₂, is considered a weak base. Using equation 6

K_a × K_b = K_w = 1.0 × 10⁻¹⁴ 

above, the K_b for

C₂H₃O₂⁻ 

is

(1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰. 

Using this value and a reaction table, the equilibrium composition of a 0.2 M

NaC₂H₃O₂ 

solution can be calculated as follows.

	C2H3O2−	+	H2O		HC2H3O2	+	OH−
initial	0.2 M				0.0 M		0.0 M
Δ	−x				+x		+x
equilibrium	0.2 − x M				x M		x M

Thus the equilibrium concentrations are

[C₂H₃O₂⁻] = 0.2 M 

and

[HC₂H₃O₂] = [OH⁻] = 1.1 × 10⁻⁵ M. 

Equation 9

Kb = [HC2H3O2][OH− ] [C2H3O2− ] = (x)(x) (0.2 − x) ≅ x2 0.2 = 5.6 × 10−10

x = 1.1 × 10−5 M

 

shows the detailed method to calculate

[OH⁻] 

for a weak base. In cases where the percent ionization of the base is less than 5%, equation 9

Kb = [HC2H3O2][OH− ] [C2H3O2− ] = (x)(x) (0.2 − x) ≅ x2 0.2 = 5.6 × 10−10

x = 1.1 × 10−5 M

 

simplifies to the expression below, where c_wb represents the original concentration of the weak base in M. The pOH of this solution can now be calculated by taking the

−log[OH⁻] 

to get 4.98. Using equation 8

pH = 14 − pOH. 

, the pH of the

0.2 M C₂H₃O₂⁻ 

solution is

14 − 4.98 = 9.02. 

When acids and bases are mixed together, things become more complicated. No matter how complicated, all mixtures of acids and bases should be approached the same way.

• 1
  Determine what species actually exist in solution. Any sodium or potassium salt will dissociate completely as will any strong acid or base.
• 2
  Write out a balanced chemical reaction to describe any interactions that will occur. Remember that the reactants must be present in the original solution as identified in step 1. Look for possible K_a or K_b reactions. Or for reactions of a weak base with a hydronium ion or a weak acid with a hydroxide ion that will react 100%. Sometimes there will not be an appropriate chemical reaction!
• 3
  Generate a reaction table based on the balanced chemical reaction. Remember that the initial concentrations for the products are not necessarily zero!
• 4
  Calculate the equilibrium composition of the mixture. This is only possible if the equilibrium constant for the reaction is known or if the reaction goes to completion.
• 5
  Calculate the pH. Once the
  [H₃O⁺] 
  or
  [OH⁻]  
  is known, the pH can be calculated as above in the pure solutions.

Figure 1 demonstrates the potential combinations that can occur when mixing acids and bases.

Mixture 1:

Mixture 1, reaction of a strong acid and strong base, goes to completion. A reaction table can be used to determine if excess acid or base remains. For example, consider the addition of 10. mL of 0.10 M HCl to 15 mL of 0.20 M NaOH. Since two solutions are being mixed, it is better to use moles or mmoles in your reaction table.

	H3O+	+	OH−	→	H2O	+	H2O
given	10. mL × 0.10 M		15 mL × 0.20 M
initial	1.0 mmol		3.0 mmol		-		-
Δ	−1.0 mmol		−1.0 mmol		-		-
final	0 mmol		2.0 mmol		-		-

In this example, only strong base remains. The

[OH⁻] 

can be calculated from the final mmol/(total volume) to give 2.0 mmol/(10 + 15 mL) = 0.08 M. Table 1 contains the necessary formulas to calculate pOH and pH from this information. The pH of this mixture is 12.90.

Mixture 2:

Mixture 2, reaction of a strong base and weak acid, also goes to completion. Depending upon the relative amounts of material, the final solution may be composed of only strong base, only weak base, or a mixture of weak acid and weak base. For example, consider the addition of 15. mL of 0.20 M NaOH to 10. mL of 0.30 M HC₂H₃O₂. Since two solutions are being mixed, it is better to use moles or mmoles in your reaction table.

	HC2H3O2	+	OH−	→	C2H3O2−	+	H2O
given	10. mL × 0.30 M		15 mL × 0.20 M
initial	3.0 mmol		3.0 mmol		-		-
Δ	−3.0 mmol		−3.0 mmol		+3.0 mmol		-
final	0 mmol		0 mmol		3.0 mmol		-

In this example, only weak base remains. The

[C₂H₃O₂⁻] 

can be calculated from the final mmol/(total volume) to give 3.0 mmol/(10 + 15 mL) = 0.120 M. Table 1 contains the necessary formulas to calculate the pH, given the concentration of weak base. The pH of this mixture is 8.91.

Mixture 3:

Mixture 3, a strong base and weak base, does not have any reaction. Rather, the

[OH⁻] 

associated with the strong base tends to suppress generation of

[OH⁻] 

from the reaction of the weak base and water. Please consult your textbook for more details. In compositions such as this, the pH can be readily obtained by focusing upon the concentration of strong base. For example, consider the addition of 20. mL of 0.20 M NaOH to 10. mL of 0.20 M NaC₂H₃O₂. This results in dilution of the strong base. In this example,

[OH⁻] 

= mmol NaOH/total volume = 20 mL × 0.20 M

OH⁻ / (20. + 10. mL) = 0.13 M. 

Table 1 contains the necessary formulas to calculate pOH and pH from this information. The pH of this mixture is 13.12.

Mixture 4:

Mixture 4, reaction of a strong acid and weak base, also goes to completion. Depending upon the relative amounts of material, the final solution may be composed of only strong acid, only weak acid, or a mixture of weak acid and weak base. For example, consider the addition of 15. mL of 0.20 M HCl to 20. mL of 0.20 M NaC₂H₃O₂. Since two solutions are being mixed, it is better to use moles or mmoles in your reaction table.

	H3O+	+	C2H3O2−	→	HC2H3O2	+	H2O
given	15. mL × 0.20 M		20 mL × 0.20 M
initial	3.0 mmol		4.0 mmol		-		-
Δ	−3.0 mmol		−3.0 mmol		+3.0 mmol		-
final	0 mmol		1.0 mmol		3.0 mmol		-

In this example, weak acid and weak base remain. This type of solution is known as a buffer and will be explored in greater detail in experiment 7. The pH can be calculated using the Henderson-Hasselbalch equation: For the current example,

pH = −log(1.8 × 10^–5) + log(1.0 mmol C₂H₃O₂⁻ / 3.0 mmol HC₂H₃O₂) = 4.26.

Mixture 5:

Mixture 5, a strong acid and weak acid, also does not have any reaction. The [H₃O⁺] associated with the strong acid tends to suppress generation of [H₃O⁺] from the reaction of the weak acid and water. In compositions such as this, the pH can be readily obtained by focusing upon the concentration of strong acid. For example, consider the addition of 30. mL of 0.20 M HCl to 20. mL of 0.15 M HC₂H₃O₂. This results in dilution of the strong acid. In this example, [H₃O⁺] = mmol HCl/total volume = 30 mL × 0.20 M H₃O⁺ / (30. + 20. mL) = 0.12 M. The pH = –log[H₃O⁺] = 0.92.

Mixture 6:

Mixture 6, the combination of a weak acid and a weak base, is also a buffer. The pH can be calculated using the Henderson-Hasselbalch equation. For example, consider the mixture of 10. mL of 0.1 M NH₄Cl and 15 mL of 0.10 M NH₃. For this example,

pH = −log(5.6 × 10^–10) + log(1.5 mmol NH₃ / 1.0 mmol NH₄⁺) = 9.43.

This is a brief summary of calculating the pH of pure acid and base solutions as well as mixtures of acid and base solutions. In this lab, you will explore these solutions in more detail and gain a greater understanding of equilibrium in aqueous acid/base mixtures.

Equipment

Reagents

Safety

HCl and NaOH are corrosive. They can attack the skin and cause permanent damage to the eyes. If either solution splashes into your eyes, use the eyewash immediately. Hold your eyes open and flush with water. If contact with skin or clothing occurs, flush the affected area with water. Have your lab partner notify your instructor about the spill.

Waste Disposal

All solutions can be flushed down the sink with plenty of water.

Prior to Class

Please enter the calculated pH values from the WebAssign pre-lab assignment in Table A on your printed worksheet.

Lab Procedure

Please print the worksheet for this lab. You will need this sheet to record your data.

In this experiment, you will be using pH electrodes connected to the MicroLab interface. pH electrodes have a thin glass bulb at the tip. They break easily and are costly to replace. Be careful not to shove the electrode into the bottom of a beaker or drop the electrode. There is a protective guard around the tip, which should remain in place at all times. The guard will not protect against careless treatment. Please use extreme care when using this equipment. Best results in using the electrodes are obtained if:

• •
  Electrodes are kept in standard pH 7 buffer solution when not in use.
• •
  Immediately prior to use, the electrodes are rinsed with deionized water and gently blotted with a tissue, then placed in the test solution.
• •
  The electrodes are rinsed and blotted again after the measurement and returned to the pH 7 buffer solution.

Part A: Calibrating the MicroLab^TM pH Electrode

Part B: Measuring the pH of Acid/Base Mixtures

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