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CHEMICAL EQUILIBRIUM - ACIDS AND BASES

There are 4 categories that acid/base problems generally fall into: strong and weak acids/bases, buffers and hydrolysis. We will examine problem samples of each type.

Strong Acids and Bases

The strength of an acid can be determined by how far to the right the equilibrium lies. This can be measured by observing the Ka of the acid. A Ka that is large indicates the presence of a strong acid; a Ka that is small, indicates a weak acid. HCl is a strong acid and has a Ka value that approaches infinity. The dissociation of HCl is practically complete and equilibrium lies completely to the right. So, the concentration of acid will equal the concentration of H3O+ (hydronium ions) produced. For example, a solution of 0.01 M HCl will completely dissociate to 0.01 M H3O+ and 0.01 M Cl-. The concentration of HCl when equilibrium has been reached will be zero. A strong base, NaOH for example, will completely dissociate. The OH- concentration will equal the strong base concentration.


Example strong acid problem: What is the pH of a 0.010 M HCl solution?

Hint: HCl is a strong acid. The H3O+ concentration will equal the HCl concentration: ([H3O+] = 0.010 M)

The pH is determined by taking the negative log of the [H3O+] concentration:

pH = -log[H3O+] = -log(0.010) = 2.00


Example strong base problem: What is the pH of a 0.010 M NaOH solution?

Hint: NaOH is a strong base. The [OH-] concentration will equal the concentration of NaOH: [OH-] = 0.010 M

pH is determined by calculating the pOH. First, take the negative log of the [OH-] concentration. Next, convert the pH to pOH.

To calculate pOH:

pOH = -log[OH-] = -log(0.010) = 2.00

The pH can now be determined by using the formula pH + pOH = 14:

pH = 14.00 - 2.00 = 12.00


Weak Acids and Weak Bases

Weak acids and bases don't dissociate completely. Equilibrium exists between a weak acid, H3O+, water, and the weak acid's anion. For all weak acids the equilibrium lies predominantly on the left, indicating that a small amount of H3O+ is produced. The Ka for a weak acid is generally a number less than 1.

You will encounter three kinds of problems dealing with weak acids and weak bases: dissociation, buffers and hydrolysis.


Dissociation of a Weak Acid

Here, you will find the H3O+ concentration and the pH of a weak acid when the initial concentration is known.

Example weak acid problem: What is the H3O+ concentration and pH of a 0.10 M solution of hypochlorous acid (HOCl)?

Hint: Ka = 3.5 x 10-8

First notice that the value of Ka is small. This means that HOCl is a weak acid. Write out the equilibrium involved:

HOCl + H2O = H3O+ + OCl-

Set Ka equal to the mass-action expression:

Create a table for initial conditions, the change in equilibrium and the final equilibrium conditions. At the beginning, the concentration of HOCl is 0.10 M. The concentrations of both H3O+ and OCl- are zero:

Note: water is not included in the calculation since it is the solvent.

For equilibrium to occur, some of the HOCl must dissociate to form H3O+ and OCl-. Since we do not know how much will dissociate, we'll label the amount of HOCl lost as -x and the amount of H3O+ and OCl- formed as +x:

Le Chatelier's principle can be used to predict the effect of a change in conditions on a chemical equilibrium. Le Chatelier's Principle states: if a stress is applied to a system in a state of dynamic equilibrium, the system will shift in a direction which opposes the stress. In the problem above, the "stress" is the absence of H3O+ and OCl- which will cause the equilibrium to shift to the right, relieving the stress by forming H3O+ and OCl-.

Using a little bit of math, you can create statements to determine the amount of each ingredient at equilibrium:

The above quantities are going be used in the mass-action expression for the equilibrium of the acid. Remember, the Ka for HOCl is 3.5 x 10-8.

The solution for x is simplified because the x in the statement 0.10 - x can be ignored. It can be ignored because its value will be extremely small compared to the concentration of 0.10 M. To determine if x can be ignored, compare the value of the last decimal place in the acid concentration to the value of the equilibrium constant. A difference greater than 100 may be ignored.

In the example problem, the concentration is 10-2 and the equilibrium constant is 10-8. The difference is 106, therefore, x can be clearly ignored. Our equation is now:

Multiply both sides by 0.10:

x2 = 3.5 x 10-9

Take the square root of both sides:

x = [H3O+] = 5.9 x 10-5 = [OCl-] (also)

Take the negative log of H3O+ to find the pH:

pH = -log[H3O+] = -log(5.9 x 10-5) = 4.23

Additionally, the HOCl concentration at equilibrium will be:

[HOCl] = 0.10 - 5.9 x 10-5 = 0.10 M (using 2 significant figures)


Example weak base problem: What is the OH- concentration and pH of a 0.10 M solution of NH3?

Hint: Kb = 1.8 x 10-5

Notice that the value of Kb is small. We are going to use the same method that we used for weak acid solutions:

We will assume that no NH4+ or OH- will have formed before equilibrium is reached.

A shift to the right is needed to establish equilibrium. The amount of NH3 that is lost(-x), and the amounts of NH4+ and OH- formed (+x) are not known, we will again assign the values of -x and +x.

Using a little bit of math, you can create statements to determine the amount of each ingredient at equilibrium:

Again, we will be using these quantities in the mass-action expression:

Determine if the x in the denominator can be ignored. The last decimal place in 0.10 is 10-2. The value of the constant is 10-5. The difference is 103. Because the difference is more than 100, the x may be ignored. Our math equation is:

Multiply both sides by 0.10:

x2 = 1.8 x 10-6

Take the square root of both sides:

x = [NH4+] = [OH-] = 1.3 x 10-3

To determine the pH, find the pOH by taking the negative log of OH-:

pOH = -log[OH-] = -log(1.3 x 10-3) = 2.87

Now subtract the pOH from 14:

pH = 14.00 - pOH = 14.00 - 2.87 = 11.13

Buffer Problems

Buffer solutions are solutions which resist change in pH upon addition of small amounts of acid or base. Buffer solutions usually consist of either a weak acid and its salt or a weak base. The resistive action is the result of the equilibrium which is set up between the weak acid and the salt. Buffers can resist pH change when a strong acid or a strong base is introduced into the solution. The easiest way to recognize a buffer problem is the concentrations of the weak acid or base and the concentrations of the conjugate acid or base are known.


Acidic Buffers

Example buffer problem: We have a solution of 0.20 M in acetic acid (HAc) and 0.10 M in sodium acetate (NaAc). What is the pH of this solution?

Set up your table:

The initial concentrations of Ac- and HAc are both known. The sodium acetate provides the acetate ion. The sodium ion will not be part of the equilibrium. It is a spectator ion and is ignored. Next, we use Le Chatelier's principle and recognize that to achieve equilibrium, some of the HAc will need to dissociate to create H3O+ ions:

Using a little bit of math, you can create statements to determine the amount of each ingredient at equilibrium:

The algebraic quantities are inserted into the mass-action expression:

Determine if the x's can be ignored. Compare them to the Ka value. In our sample problem, the concentrations of the weak acid and its conjugate base are 10-2 and the Ka is 10-5. The difference is greater than 100 so both of the x quantities may be ignored. Our algebraic expression is now:

Solve for x to determine the H3O+ ion concentration:

Take the negative log of the H3O+ ion concentration to find the pH:

pH = -log[H3O+] = -log[3.6 x 10-5] = 4.44


Alkaline Buffers

Example buffer problem: We have an alkaline buffer of 0.20 M aqueous ammonia (NH3) and 0.10 M ammonium chloride (NH4Cl). What is the pH of the solution?

NH4+ is a common ion to the ammonia equilibrium. The chloride ion does not participate in the equilibrium and will be a spectator ion which will be ignored.

Create your table:

The concentrations are known for the weak base and conjugate acid. Applying Le Chatelier's Principle you'll notice that the equilibrium shifts to the right creating OH-. Some NH3 will be lost and some NH4+ will be formed as well as some OH-.

Fill in your chart for the amount of NH3, NH4+ and OH- at equilibrium:

Insert the above into the mass-action expression:

Determine if the x's can be ignored. Compare them to the Kb value. Here, the concentrations 10-2 and the Kb are known to the 10-5 magnitude. The difference is greater than 100, the x quantities may be ignored. We now have:

Solving for x will give us the concentration of the hydroxide ion:

Find the pOH (take the negative log of the OH- ion concentration):

pOH = -log[OH-] = -log(3.6 x 10-5) = 4.44

Subtract the pOH from 14.00 to determine pH:

pH = 14.00 - pOH = 14.00 - 4.44 = 9.56

Hydrolysis Problems

Hydrolysis involves salts of weak acids and bases. Salt is a term used for ionic compounds composed of positively charged cations and negatively charged anions. Cations of salt come from the base and anions come from the acid. An easy way to remember this is to keep the vowels (Anions-Acids) and consonants (Cations-Bases) together. Not all salts are neutral. Some salts hydrolyze in water and will result in either an acidic solution or a basic solution. Sodium Chloride, NaCl, will form a neutral solution meaning, no hydrolysis occurs.

The sodium ion (cation), which came from NaOH (a strong base), is a weak conjugate acid which will not hydrolyze. It will act as a spectator ion. Also, the chloride ion (anion), which came from HCl (a strong acid), is a weak conjugate base which will not hydrolyze. It will also act as a spectator ion. Therefore, the only H3O+ ions in the solution are a result of the self-ionization of water, and the solution will have a neutral pH.


Example hydrolysis problem: Sodium acetate with a concentration of 0.10 M:

The sodium ion (cation), came from NaOH (a strong base), and will not hydrolyze. The acetate ion (anion), came from acetic acid (a weak acid). The acetate ion is a strong conjugate base and will hydrolyze. It will behave as a Lowry-Bronsted base (A theory of acids and bases which considers acids as proton donors and bases as proton acceptors), and will accept a proton from water. See the equilibrium below:

C2H3O2- + H2O = HC2H3O2 + OH-

The OH- ion is made and the pH of the salt solution is alkaline. To determine the solution's pH we need the acetate ion's initial concentration and the value of Kb, the equilibrium constant (this is sometimes denoted as Kh, hydrolysis constant). The equilibrium constant is calculated using the Kw of water and the Ka of the weak acid:

The concentration of sodium acetate was given as 0.10 M. The concentration of acetate ion is 0.10 M as well because for every mole of sodium acetate there is also 1 mole of acetate ions. You can calculate the pH just as we have for the previous equilibrium problems. The first step is to create your chart:

To establish equilibrium, the reaction is going to shift to the right as there is no hydroxide or acetic acid present yet (Le Chatelier's Principle). An x amount of acetate ion (-x) will be consumed. An x amount of acetic acid and hydroxide ion (+x) will be formed:

Using a little bit of math, you can create statements to determine the amount of each ingredient at equilibrium:

Substitute the quantities into the mass-action expression:

Determine if the "x" in the denominator can be ignored. The initial concentration of the acetate ion is (10-2) and the equilibrium constant is (10-10). The difference is 108.

Multiply both sides by 0.10:

x2 = 5.6 x 10-11

Take the square root of both sides:

x = 7.5 x 10-6 = [OH-]

x is equal to the concentration of the OH- ion and the pH can now be calculated. First, determine the pOH:

pOH = -log[OH-] = -log(7.5 x 10-6) = 5.12

Subtract the pOH value from 14.00 to determine pH:

pH = 14.00 - pOH = 14.00 - 5.12 = 8.88

The solution is alkaline.


Example hydrolysis problem: A 0.10 M solution of NH4Cl. What is the pH of the solution? Determine if hydrolysis will occur.

The chloride anion (a weak conjugate base) of HCl (a strong acid) will not hydrolyze. The ammonium cation (a strong conjugate acid) of NH3 (a weak base) will hydrolyze. It will donate a proton to water (Lowry-Bronsted acid):

NH4+ + H2O = NH3 + H3O+

Because hydronium ion is created, the solution will be acidic. To determine the solution's pH, we need the salt's initial concentration as well as the equilibrium constant. The equilibrium constant is calculated using the Kw of water and the Kb of the base:

Set up your table using an 0.10 M as the initial concentration of salt:

We will assume that no ammonia or hydronium ion will be produced before equilibrium has been reached. To achieve equilibrium, an x amount of ammonium ion (-x) will be lost. An x amount of ammonia and hydronium ion (+x) will be formed. To achieve equilibrium, the reaction will shift to the right (Le Chatelier's Principle). Enter the changes in concentration to your chart:

Using a little bit of math, you can create statements to determine the amount of each ingredient at equilibrium:

Substitute the quantities into the mass-action expression:

Determine if the "x" in the denominator can be ignored. Use the initial concentration 10-2 and the equilibrium constant 10-10. The difference is 108.:

Multiply both sides by 0.10:

x2 = 5.6 x 10-11

Take the square root of both sides:

x = [H3O+] = 7.5 x 10-6

x is equal to the hydronium ion concentration. Calculate the pH by taking the negative log of the hydronium ion concentration:

pH = -log[H3O+] = -log(7.5 x 10-6) = 5.12

The solution is acidic.

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