Laboratory 11.1: Determine the Effect of Concentration on pH

This article incorporates, in modified form, material from Illustrated Guide to Home Chemistry Experiments: All Lab, No Lecture.

 

pH is a metric used to specify the acidity (or basicity, also called alkalinity) of an aqueous solution. The pH of a solution is determined by the relative activities of the hydronium (H₃O⁺)ions and the hydroxide (OH^–)ions in the solution. The pH of a solution in which the activities of these two ions are equal – such as pure water at 25 ºC – is 7.00. A solution in which the activity of the hydronium ions is higher than the activity of the hydroxide ions has a pH lower than 7 and is acidic. A solution in which the activity of the hydroxide ions is higher than the activity of the hydronium ions has a pH greater than 7, and is basic.

 

Activity versus Concentration

Technically, the pH of a solution depends on the activity of the hydronium ions rather than their concentration, but it’s easy to measure concentration accurately and very difficult to measure activity accurately. (Activity can be thought of as the effective concentration of hydronium ions, which is lower than the actual concentration because in solution some of the hydronium ions are “screened” by other ions present in the solution and therefore unable to participate in reactions.) In very concentrated solutions, activity may be considerably lower than the actual concentration. In dilute solutions, concentration and activity correlate very closely, so the concentration is typically used for calculations.

pH is specified on a log₁₀, scale, which allows a very wide range of activities (concentrations) to be specified using a small range of numbers. A difference of one pH number corresponds to a difference of ten times in acidity or basicity. For example, a solution with a pH of 5 is ten times (10¹) more acidic than a solution with a pH of 6, and a solution with a pH of 9 is ten times (10¹) more basic than a solution with a pH of 8. Similarly, a solution with a pH of 2 is 10,000 (10⁴) times more acidic than a solution with a pH of 6, and a solution with a pH of 12 is 10,000 (10₄) times more basic than a solution with a pH of 8. Although the range of pH values is usually considered to be 0 through 14, an extremely acidic solution (such as a concentrated solution of hydrochloric acid) can have a pH lower than 0, and an extremely basic solution (such as a concentrated solution of sodium hydroxide) can have a pH greater than 14.

For relatively dilute solutions of strong acids and bases, you can estimate pH using the formula:

pH = -log₁₀[H₃O⁺]

where [H₃O⁺is the concentration of the hydronium ion in mol/L. For example, hydrochloric acid dissolves in water according to the following equation:

HCI + H₂O→H₃O⁺+CI^–

Because HCl is a strong acid, the reaction proceeds to completion, which is to say that essentially all of the HCl reacts to form hydronium ions and chloride ions. The approximate pH of a 0.01 M solution of hydrochloric acid is:

pH = -log₁₀[0.01] = 2

In calculating that approximate pH, we assume that the hydrochloric acid fully dissociates in solution into H₃O⁺ ions and CI^–. For strong acids like hydrochloric acid, that’s a reasonable assumption. For weak acids, such as acetic acid, that assumption is not valid, because weak acids dissociate only partially in solution. The concentration of hydronium ions in a solution of a weak acid is lower (perhaps much lower) than the concentration of the acid itself.

When acetic acid dissolves in water, the dissociation reaction looks like this:

CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO^–

This reaction reaches an equilibrium, with reactants being converted to products and vice versa at the same rate. Therefore, in a 1.0 M solution of acetic acid (about the concentration of household vinegar), the actual concentration of the hydronium ion is something less than 1.0 M, because some of the acetic acid remains undissociated. Based on the previous calculation, we know that the pH of a 1.0 M solution would be about 0 if the acid had fully dissociated, so we know that the actual pH of the 1.0 M acetic acid will have some value higher than 0.

If we assume for a moment that at equilibrium only 10% of the acetic acid has dissociated, we can calculate the approximate pH of the solution. If only 10% of the acid has dissociated in a 1.0 M solution of acetic acid, the hydronium ion concentration is 0.1 M. Filling in the formula, we get:

pH = -log₁₀[0.1] = 1

But to estimate the pH of this solution accurately, we need a better value than our guesstimate of 10% dissociation. We get that value by looking up the equilibrium constant for the dissociation reaction shown above. In the context of pH, the equilibrium constant is referred to as the acidity constant, acid dissociation constant, or acid ionization constant, and is abbreviated K_a.

K_a versus pK_a

The acidity constant may be specified directly as K_a or as the negative logarithm of K_a, which is abbreviated as pK_a. For example, at 25 ºC, the K_a of acetic acid is 1.74·10^-5 and the pK_a is 4.76. Either value can be used by adjusting your calculations accordingly.

Paul Jones comments

I’d say pK_a is used far more today than K_a. Also, you might mention that any compound with hydrogen can, plausibly, be an acid. However, if the pKa of the compound is above 14 it essentially won’t dissociate at all in water. Also, when pH = pK_a, the acid is 50% dissociated; another good reason to use pKa.

With the acid dissociation constant for acetic acid known to be 1.74·10^-5, and ignoring the tiny contribution to [H₃O⁺] made by the water, we can calculate the pH of a 1.0 M solution of acetic acid as follows:

K_a = 1.74·10^-5 = ([H₃O⁺]·[CH₃COO^–]/[CH₃COOH])

An unknown amount of the acetic acid has dissociated, which we’ll call x. That means that the concentration of the acetic acid, or [CH₃COOH] is (1.0 – x), while the concentrations of the dissociated ions, [H₃O⁺] and [CH₃COO^–], are both x. Filling in the formula gives us:

1.74·10^-5 = ([x]·[x]/[1.0 – x])

or

1.74·10^-5 = x²/(1.0 – x)

or

(1.74·10^-5)·(1.0 -x) = x²

or

1.74·10^-5 – (1.74·10^-5·x) = x²

Solving for x gives us a value of about 4.17·10^-3 or about 0.00417. Our original guesstimate that 10% of the acetic acid would dissociate was wildly high. In fact, only about 0.417% of the acetic acid dissociates. Knowing that value tells us that the concentration of the hydronium ion in a 1 M acetic acid solution is 0.00417 M. Plugging that value into the formula allows us to calculate the approximate pH of the 1.0 M acetic acid solution:

pH = -log₁₀[0.00417] = 2.38

Although we’ve focused until now on acids rather than bases, remember that the concentrations of the hydronium ion and the hydroxide ion are related. We know that pure water at 25 °C has a pH of 7.00 and that the concentrations of hydronium ions and hydroxide ions are equal. A pH of 7.00 tells us that the concentration of hydronium ions, [H₃O⁺], is 1.00·10^-7, which means that the concentration of hydroxide ions, [OH^–], must also be 1.00·10^-7. We know that hydronium ions and hydroxide ions react to form water:

H₃O⁺ + OH^– ⇌ H₂O

According to Le Chatelier’s Principle (see Chapter 13), in a system at equilibrium, increasing the concentration of one reactant forces the reaction to the right, producing more product. Increasing [H₃O⁺] reduces [OH^–] proportionately, and vice versa. Expressed as a formula, the equilibrium constant is:

K = [H₃O⁺]·[OH^–]

or

K = (1.00·10^-7)·(1.00·10^-7) = 1.00·10^-14

In other words, the product of the concentrations of the hydronium ions and the hydroxide ions always equals 1.00·10^-14. If you increase the concentration of hydronium ions by a factor of 10 (or 10,000), the concentration of hydroxide ions decreases by a factor of 10 (or 10,000), and vice versa.

In this lab, we’ll determine the pH of solutions of two strong acids, a weak acid, a strong base, and a weak base at various concentrations.

Required Equipment and Supplies

• goggles, gloves, and protective clothing
• balance and weighing paper
• beaker, 150 mL (6)
• volumetric flask, 100 mL
• pipette, 10 mL
• pH meter
• hydrochloric acid, 1 M (100 mL)
• sulfuric acid, 1 M (100 mL)
• acetic acid, 1 M (100 mL)
• sodium hydroxide, 1 M (100 mL)
• sodium carbonate, 1 M (100 mL)
• distilled or deionized water (boil and cool before use)

All of the specialty lab equipment and chemicals needed for this and other
lab sessions are available individually from Maker Shed or other laboratory
supplies vendors. Maker Shed also offers customized laboratory kits at special
prices, including the Basic Laboratory Equipment Kit, the Laboratory Hardware Kit, the Volumetric Glassware Kit, the Core Chemicals Kit, and the
Supplemental Chemicals Kit.

Substitutions and Modifications

• You may substitute foam cups or similar containers for the beakers.
• You may substitute a 100 mL graduated cylinder for the 100 mL volumetric flask and a 10 mL graduated cylinder for the 10 mL pipette, with some loss of accuracy. (Unless your pH meter is very accurate, instrumental error will be greater than volumetric error anyway.)
• You may substitute narrow-range pH testing paper for the pH meter, with some loss of accuracy. (Typical inexpensive pH meters are accurate to 0.1 or 0.2 pH, while typical narrow-range pH testing paper is accurate to 0.4 or 0.5 pH.)
• You may reduce the quantities of the acid and base solutions, depending on the size and type of probe used by your pH meter, and use smaller containers. If your meter has a very thin probe (or if you are using pH paper) you can reduce the quantities from 100 mL to 10 mL and use test tubes.
• If you don’t have a 1 M bench solution of hydrochloric acid available, you can make it up by diluting 8.33 mL of 12 M (37%) concentrated hydrochloric acid to 100 mL. If you have hardware-store muriatic acid that lists the contents as 31.45% HCl, you can make up a 1 M solution by diluting 9.71 mL of acid to 100 mL.
• If you don’t have a 1 M bench solution of sulfuric acid available, you can make it up by diluting 5.56 mL of 18 M (98%) concentrated sulfuric acid to 100 mL. If you have hardware-store battery acid that lists the contents as 35% sulfuric acid, you can make up a 1 M solution by diluting 15.57 mL of acid to 100 mL.
• If you don’t have a 1 M bench solution of acetic acid available, you can make it up by diluting 5.75 mL of 17.4 M (99.8%) concentrated (glacial) acetic acid to 100 mL. Alternatively, most distilled white vinegar contains 5% to 6% acetic acid, which is close enough to 1 M to use for this experiment.
• If you don’t have a 1 M bench solution of sodium hydroxide available, you can make it up by dissolving 4.00 g of sodium hydroxide in water and making up the solution to 100 mL.
• If you don’t have a 1 M bench solution of sodium carbonate available, you can make it up by dissolving 10.60 g of anhydrous sodium carbonate in water and making up the solution to 100 mL.
• I suggest boiling and cooling the distilled or deionized water before use to eliminate any dissolved carbon dioxide gas, which forms carbonic acid and lowers the pH below the 7.00 pH of pure water. Alternatively, you can use distilled or deionized water from a freshly-opened container or one that has been kept tightly capped.

Procedure

1. If you have not already done so, put on your splash goggles, gloves, and protective clothing.
2. Label six beakers #1 through #6.
3. Pour about 100 mL of 1 M hydrochloric acid into beaker #1.
4. Using the 10 mL pipette, transfer 10.00 mL of the 1 M hydrochloric acid to the 100 mL volumetric flask.
5. Fill the volumetric flask to the reference line with distilled water, mix thoroughly, and transfer the 0.1 M solution of hydrochloric acid to beaker #2.
6. Proceeding from beaker to beaker, repeat steps 4 and 5 until you have 0.01 M, 0.001 M, 0.0001 M, and 0.00001 M solutions of hydrochloric acid in beakers #3 through #6, respectively.
7. Read and follow the directions for your pH meter with respect to calibrating it, rinsing the electrode between measurements and so on. Use the pH meter to measure the pH of the solutions in beakers #1 through #6, and record those values on Line A of Table 11-1.
8. Repeat steps 3 through 7 for the solutions of sulfuric acid, acetic acid, sodium hydroxide, and sodium carbonate.
9. Based on your observations, calculate the pK_avalues for the acids and the pK_bvalues for the bases, and enter your calculated values in Table 11-1.

Figure 11-1. Using a pH meter to determine the pH of a solution

Table 11-1. Determine the effect of concentration on pH – observed and calculated data

Solute	Beaker #1 (1.0 M)	Beaker #2 (0.1 M)	Beaker #3 (0.01 M)	Beaker #4 (0.001 M)	Beaker #5 (0.0001 M)	Beaker #6 (0.00001 M)	pK
A. hydrochloric acid	___.___pH	___.___pH	___.___pH	___.___pH	___.___pH	___.___pH	___.___
B. sulfuric acid	___.___pH	___.___pH	___.___pH	___.___pH	___.___pH	___.___pH	___.___
C. acetic acid	___.___pH	___.___pH	___.___pH	___.___pH	___.___pH	___.___pH	___.___
D. sodium hydroxide	___.___pH	___.___pH	___.___pH	___.___pH	___.___pH	___.___pH	___.___
E. sodium carbonate	___.___pH	___.___pH	___.___pH	___.___pH	___.___pH	___.___pH	___.___

Disposal

Retain the 1.0 M acid and sodium hydroxide solutions for use in the following lab. Neutralize the other solutions, beginning with the most dilute samples, by pouring them into a bucket or similar container. The contents of the bucket can be flushed down the drain.

Review Questions

Q1: Chart pH against concentration for all five compounds on one piece of graph paper. What do you conclude about the effect of concentration for a strong acid (hydrochloric acid or sulfuric acid) versus a weak acid (acetic acid) and for a strong base (sodium hydroxide) versus a weak base (sodium carbonate)?

Q2: What pH values would you expect for 10 M solutions of hydrochloric acid and acetic acid? Why?

Q3: If acids A and B have pKa values of -9.32 and 2.74, respectively, which is the stronger acid? Why?

Q4: The [H3O+] of a solution is known to be 0.000413 mol/L. What is the pH of the solution?

Q5: The [OH-] of a solution is known to be 0.000413 mol/L. What is the pH of the solution?

Q6: A 0.1 M solution of an acid is found to have a pH of 2.37. What is the pKa of that acid?

Q7: Knowing only the molarity and pH of an unknown acid solution, how might you identify the acid?