CHAPTER 17 ACID – BASE EQUILIBRIA. I. INTRODUCTION A) Acid strength is measured by the extent of the overall reaction of the acid with water. 1) Strong.

CHAPTER 17 ACID – BASE EQUILIBRIA

I. INTRODUCTION A) Acid strength is measured by the extent of the overall reaction of the acid with water. 1) Strong acids go 100% to the right. HCl + H 2 O  H 3 O + + Cl -

2) Weak acids go "slightly" to the right. HA + H 2 O  H 3 O + + A - Why can we make [H 2 O] K c = to K a ?

K a is known as the acid dissociation constant and the acid ionization constant. CH 3 COOH + H 2 O  H 3 O + + CH 3 COO -

3) Some K a 's at 25 o C: 4) Some questions I will want you to be able to answer from these data are: a) Which of the above acids is the strongest? b) Which is the weakest acid? c) What is the strongest conjugate base?

II. HOW CAN WE FIND K a 's? A) The most common way to obtain a K a is to measure the pH of a solution prepared by dissolving a known amount of weak acid to form a given volume of solution. B) An example: The pH of a 0.10 M solution of HOCl is 4.23. What is the value of K a ?

How do we get the values of [H 3 O + ], [OCl - ] and [HOCl]? From the pH we can obtain what value? 4.23 = - log [H 3 O + ] - 4.23 = log [H 3 O + ] 5.9 X 10 -5 = [H 3 O + ] Then we know the value of the [OCl - ] = ? The value of [HOCl] = ?

We set up a table like we did in Chapter 15.

Then we put the values in the appropriate equation for K a.

III. DEGREE OF DISSOCIATION AND PERCENT DISSOCIATION (OR IONIZATION). A) The degree of dissociation is the fraction of molecules that react with water to give ions.

If the [H + ] in a 0.60 M solution of HF is 0.021 M, what is the degree of dissociation? % ionization is obtained by multiplying the degree of dissociation by 100 and adding the % sign.

C) The larger the K a the greater the degree of dissociation (the greater the % dissociation). The stronger the acid the greater the % ionization. D) As the original concentration decreases the percent dissociation increases. How do we explain this situation?

IV. Calculation of the Concentration of Species in a Weak Acid Solution Using K a - THE APPROXIMATE METHOD

A) What are the concentrations of all species in a 0.10 M of acetic acid (CH 3 COOH)? What is the pH of the solution? K a for acetic acid is 1.7 X 10 -5. You will need a balanced equation and a chart. CH 3 COOH + H 2 O  H 3 O + + CH 3 COO -

Then we put the results of the chart into the equation for K a.

To avoid using the quadratic equation and focus more attention on what is happening chemically, we THINK AS FOLLOWS: Since K a is so small, x can be ASSUMED to be much less than 0.10. That makes 0.10 - x = 0.10. Then we don't have to use the quadratic equation to obtain x.

The above expression for Ka becomes:

BUT YOU will have to check your assumption: To check if the assumption is ok, we will use the 5% rule as given in the text. How do we calculate the 5%? Another check to see if the assumption is ok is to check to see if the precision rule allows you drop the x. In the above case: 0.10 - 0.0013 = 0.10 (The answer can only be reported to hundredths.)

The answer to the question then includes the following: [CH 3 COOH] = 0.10 M [H 2 O] = 55.5 M [H 3 O + ] = 1.3 X 10 -3 M [CH 3 COO - ] = 1.3 X 10 -3 M And there is one more species in the solution. What is it and what is its concentration?

To answer the question, we still have to calculate the pH. pH = - log 1.3 x 10 -3 = 2.89

B) Calculate the [H 3 O + ] in a 0.100 M solution of nitrous acid, HNO 2, for which the K a is 4.5 X 10 -4. You will need an equation and a chart. HNO 2 + +H 2 O  H 3 O + + NO 2 -

Now we have to see if the assumption is ok!! 0.100 - 0.0067 gives an answer to three digits (0.093) so that indicates the approximation is not appropriate for this problem.

Also, IF YOU TAKE: You cannot use the approximate method, you must use the quadratic equation. What answer do we obtain with the quadratic equation?

Work some extra credit problems. C) Calculating concentrations of species in a solution of a diprotic acid. (What is a diprotic acid? 1) We will examine the stepwise ionization of oxalic acid, a diprotic acid found in spinach and rhubarb.

H 2 C 2 O 4 + H 2 O  H 3 O + + HC 2 O 4 - K a1 = 5.90 x 10 -2 HC 2 O 4 - + H 2 O  H 3 O + + C 2 O 4 2- K a2 = 6.40 x 10 -5 Now what do we do with this information? If the K a 's are far apart, we can make the problem easier. We first compare the K a 's to see which one is more important in delivering H 3 O + 's to the solution.

In this problem K a1 is much greater than K a2. We will assume that all the H 3 O + 's came from the first equation, when calculating the [H 3 O + ], [HC 2 O 4 - ], and the [C 2 O 4 2- ] in the 0.10 M H 2 C 2 O 4 solution. You will need an equation and a chart.

H 2 C 2 O 4 + H 2 O  H 3 O + + HC 2 O 4 - K a1 = 5.90 x 10 -2 We now put these values in the general equation:

K a is fairly large so the approximate method cannot be used. You must use the quadratic equation to obtain the value of x. The above K a relation becomes: x 2 + 5.90 X 10 -2 x - 5.9 X 10 -3 = 0 The positive root gives x = _____ = [H 3 O + ] = [HC 2 O 4 - ] We then proceed to the second ionization, where we make a second chart corresponding to the second reaction.

HC 2 O 4 - + H 2 O  H 3 O + + C 2 O 4 2- K a2 = 6.40 x 10 -5 We then place these values in the equation for K a2.

Since K a2 is small we can assume that y << 0.053. When we do this the above equation becomes:

We obtain the following values for the concentrations of the following species in solution: [H 3 O + ] = 0.053 M [HC 2 O 4 - ] = 0.053 M [C 2 O 4 2- ] = 6.4 X 10 -4 M [H 2 O] = 55.5 M [H 2 C 2 O 4 ] = 0.10 - 0.053 = 0.05 M [OH - ] = 1.00 X 10 -14 divided by 0.053 is 1.9 X 10 -13 M

V. Base Ionization Equilibrium A) Weak bases have a K b. NH 3 + H 2 O  NH 4 + + OH - B) Some other weak bases are what can be called organic relatives of ammonia.

CH 3 NH 2 + H 2 O  CH 3 NH 3 + + OH - Kb = 4.4 X 10 -4 (CH 3 ) 2 NH + H 2 O  (CH 3 ) 2 NH 2+ + OH - K b = 5.1 X 10 -4 (CH 3 ) 3 N + H 2 O  (CH 3 ) 3 NH + + OH - K b = 7.4 X 10 -5 C 6 H 5 NH 2 + H 2 O  C 6 H 5 NH 3 + + OH - K b = 4.2 X 10 -10

C) Calculate the pH of a 1.0 M solution of methylamine, CH 3 NH 2. What will be your guess at the answer? _____ We need the equation, a chart and the set- up for K b. CH 3 NH 2 + H 2 O  CH 3 NH 3 + + OH -

We then assume that x << 1.0 and we obtain the following expression:

We then obtain the [OH - ] = [CH 3 NH 3 + ] = 2.1 X 10 -2 M. We have to check the assumption: [OH - ] = 2.1 X 10 -2 M pOH = - log [OH - ] = - log 2.1 X 10 -2 = 1.68 This is not the answer to the question. Remember the question is about pH, not pOH.

So the pH = 14.00 - 1.68 = 12.32 Some extra credit problems for you to do in class. VI. Acid and Base Properties of Salts A) I want you to be able to predict whether the solution of a salt is acidic, basic or neutral. B) How will you be able to do this?

C) You will have to recall the 6 strong acids and 6 strong bases, and call all others are WEAK. 1) Salts which have cations from strong bases and the anions of strong acids have no effect upon the [H + ] when dissolved in water. EXAMPLE: NaCl

Na + is the cation of NaOH, a very strong base, therefore Na + is a very weak conjugate acid. It does not react with water to take an OH - from it. Cl - is the anion of HCl, a very strong acid, therefore Cl - is a very weak conjugate base. It does not react with water to take an H + from it. We expect that the water solution of NaCl to be neutral.

Na + + H 2 O ---> N.R. Cl - + H 2 O ---> N.R. Neutral solution results. We have not changed the concentration of H 3 O + or the concentration of the OH - ions by inserting NaCl into the water.

2) Salts which have cations from strong bases and the anions of weak acids produce a basic solution when dissolved in water. EXAMPLES: NaC 2 H 3 O 2 sodium acetate NaC 2 H 3 O 2 in water dissociates 100% to Na + ions and CH 3 COO - ions. Na + is the cation of NaOH, a very strong base, therefore Na + is a very ________ ________ _______________. It does not react with water to take an OH from it.

CH 3 COO - is the anion of the weak acid, acetic acid, therefore its conjugate base is stronger and will react with water to produce more OH- ions. Na + + H 2 O ----> N.R. CH 3 COO - + H 2 O  CH 3 COOH + OH - Do you see how a basic solution results from the action of the conjugate base with water?

Another example is NaF. What happens with it? It dissociates 100% into ________ions and ________ions. Which of these ions can react with water? What kind of solution will it make? Will the pH of the solution be greater or less than 7?

3) A salt which has the cation from a weak base and the anion of a strong acid produce an acidic solution. EXAMPLE: NH 4 Cl NH 4 Cl ----> NH 4 + + Cl - Cl- + H 2 O ----> N.R. (Why?) NH 4 + + H 2 O  NH 3 + H 3 O + We see that the concentration of the H 3 O + increases so the pH of the solution will be __________________ 7.

4) Salts derived from a weak acid and a weak base have both ions reacting with water (undergoing hydrolysis). a) Whether the solution is acidic or basic depends on the relative acid - base strengths of the two ions. To determine this, you will need to compare the cation's K a with the K b of the anion.

b) If the K b of the anion > the K a of the cation, the solution is basic. c) If the K b of the anion < the K a of the cation, the solution is acidic. d) If the K b of the anion = the K a of the cation, the solution is neutral. B) How do we obtain K a 's and K b 's for ions?

What is the K b for the CH 3 COO - ion? We start with CH 3 COONa. It dissociates 100% CH 3 COONa ----> CH 3 COO - + Na + Na + + H 2 O ----> N.R. CH 3 COO - + H 2 O  CH 3 COOH + OH -

Generally the values for the K a 's and K b 's for ions are not given in books. How can we obtain them if this is the case? What will be given in the text and how can we use it? The K a for the weak acid will be given. How is that related to the K b for the conjugate base?

Compare K a with K b.

If we look closely, we see that part of K b is upside down from the K a. So let's invert it. BUT this still isn't equal to K b. We have to eliminate H 3 O + from the denominator and get OH - in the numerator. How can we do this?

So for the acetate ion, the value of K b is obtained by the following division: This is a small number, but large enough to affect the pH of the solution. We can conclude from this that K w = K a K b

What is the pH of a 0.10 M solution of CH 3 COONa? CH 3 COONa ----> CH 3 COO - + Na + Na + + H 2 O ----> N.R. CH 3 COO - + H 2 O  CH 3 COOH + OH - We have the equation, now we need a chart.

To solve for x we put these values in the equation for K b for the ion.

That is not the answer yet, remember the question asks for the pH of the solution. So we have a further calculation to do. Notice that we are very close to having to use the 1 X 10 -7 to add to the [OH - ] ions. For this class we won't do it, but for a more precise answer you would put it in.

C) What is the K a for the NH 4 + ion? The salt of NH 4 + with non-complicating factors is NH 4 Cl (or Br -, I -, NO 3 - ). We need the equation for the reaction of NH 4 + with water. NH 4 + + H 2 O  NH 3 + H 3 O + Ka for the anion is going to equal what?

Now that we know the K a for the NH 4 + we should be able to calculate the pH of a 0.10 M solution of NH 4 Cl.

What kind of solutions (acid or base) will we get if we dissolve CuCl 2 in water; AlCl 3 ? Show by appropriate reactions of the hydrolysis of the reacting ions.

VII. THE COMMON ION EFFECT A) The Common Ion Effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion that takes part in the equilibrium.

1) Calculate the pH of a 0.50 M acetic acid solution. CH 3 COOH + H 2 O  H 3 O + + CH 3 COO -

2) What is the pH of a solution which is 0.50 M in CH 3 COOH and 0.50 M in CH 3 COONa? The most important thing that needs to come to mind when you see two similar but different compounds is to write TWO separate equations and draw two separate charts. Do not ever combine the two reactants in one equation. Crap results…

CH 3 COONa ’ CH 3 COO - + Na + CH 3 COOH + H 2 O  H 3 O + + CH 3 COO -

Now we have to put the resultant values in the expression for K a. If x is << 0.50 then we obtain the following:

The pH has gone from 2.54 to 4.77 [H 3 O + ] has gone from 2.9 X 10 -3 to 1.7 x 10 -5 Is this reasonable taking into account all that we have studied? What about Le Chatelier's Principle?

VIII. BUFFERS A) A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. B) The most important application of the common ion effect is for making buffer solutions.

C) Aqueous solutions of the body have a characteristic pH. The blood, for example, has a pH between 7.3 and 7.5. Death generally results at pH's below 7.0 and above 7.9. The blood's ability to control pH is remarkable in that many body reactions produce acids. When these enter the blood, if the blood were not buffered, its pH would be changed drastically with any body function, and death would result.

D) Some important buffer systems in the blood include the H 2 CO 3 /HCO 3 - (from CO 2 dissolving in the blood) and H 2 PO 4 - / HPO 4 2-. E) Buffers contain either a weak acid and its conjugate base or a weak _____ and its________ _______. The acid and its salt or the _______and its ______are in water solution. By choosing the correct mixture of substances we can make buffers of selected pHs.

F) Let's look at our common ion solution prepared previously, and add a small amount of HCl to the solution. CH 3 COOH + H 2 O  H 3 O + + CH 3 COO - To make the arithmetic easier, we'll assume that initially all the HCl reacts with the CH 3 COO - to form CH 3 COOH molecules.

Then, of course, some of the molecules react in the forward direction:

We now put these values in the expression for K a. pH = 4.75 compared to 4.77 before the HCl was added. Why did the pH go down?

NOW, I want you to compare this to water only. What is the pH of pure water? What is the pH of a 0.01 M water solution of HCl? How many units does the pH change? How many times is this? G) Biologists and biochemists often calculate the pH of a buffer solution using the Henderson-Hasselbach Equation.

A buffer is made containing 0.10 M CH 3 COOH ( pK a = 4.77 ) and 0.050 M CH 3 COONa. What is the pH of this solution?

H) In lab you were asked to make a buffer solution of a certain pH knowing K a. Given acid and conjugate base solutions of the same concentrations, the ratio of the concentrations becomes the ratio of the _______________ of the two samples.

1) Calculate the ratio of concentrations of HOCl (hypochlorous acid) and NaOCl (sodium hypochlorite) needed to produce a buffer solution with a pH of 7.60. K a for HOCl = 2.9 X 10 -8 HOCl + H 2 O  H 3 O + + OCl -

Translated into volumes, what does this ratio mean?

IX. TITRATIONS A) There are three types: 1) strong acid and strong base 2) weak acid and strong base 3) strong acid and weak base B) Those involving a strong acid and a strong base

1) Example: titrate a 25.00 mL of 0.100 M HCl with a 0.100 M NaOH solution. The NaOH solution is added from the buret to the HCl in the flask. If we monitor the titration with a pH meter, we obtain the following results:

After 25.00 mL of base have been added, we call that the equivalence point, because a stoichiometric amount of the reactant has been added. For the strong acid-strong base titration, the pH at the equivalence point is _____. A neutral solution has been produced.

How can we calculate the pH of a solution after a certain amount of NaOH has been added to the HCl solution? Example: What is the pH of a solution after 10.0 mL of 0.100 M NaOH have been added to 25.0 mL of 0.100 M HCl?

C) Weak acid and strong base titration Why isn't the equivalence point at pH = 7?

D) What would the titration curve of a titration of strong acid with a weak base look like? At about what pH would its equivalence point lie?

CHAPTER 17 ACID – BASE EQUILIBRIA. I. INTRODUCTION A) Acid strength is measured by the extent of the overall reaction of the acid with water. 1) Strong.

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CHAPTER 17 ACID – BASE EQUILIBRIA. I. INTRODUCTION A) Acid strength is measured by the extent of the overall reaction of the acid with water. 1) Strong.

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Presentation on theme: "CHAPTER 17 ACID – BASE EQUILIBRIA. I. INTRODUCTION A) Acid strength is measured by the extent of the overall reaction of the acid with water. 1) Strong."— Presentation transcript:

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