How to solve Acid Base Chemistry and Buffers Problem ?

Solving Buffer Problems   Buffer problems can be quite challenging for many students.  It is the purpose of this handout to help guide you through simple buffer problems you might encounter in this course.  The best way to solve these problems, and many other problems for that matter, is to …

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Solving Buffer Problems

 

Buffer problems can be quite challenging for many students.  It is the purpose of this handout to help guide you through simple buffer problems you might encounter in this course.  The best way to solve these problems, and many other problems for that matter, is to follow a step-by-step pattern. I will show you the pattern needed for buffer problems in this document.

When making a buffer, one must consider two major concepts: First, what chemical species are going to actually release or absorb added hydrogen ions; and second, at what concentrations do these species need to be in order to be an effective buffer.  Buffers are not generally effective more than 2 pH units away from the pKa of the weak acid/base pair of the buffer, and normally buffers are made using ingredients whose pKa’s are as close as possible to the desired pH of the buffer.  Additionally, the more concentrated the buffer, the better able it is to resist a change in pH.  However, considerations of ionic strength often preclude the use of concentrated buffering systems.  A compromise involving pH, pK, and concentration is often necessary.

Consider the following question:[wp_campaign_1]

Starting with 1.0M acetic acid and 0.50M NaOH, how would you prepare 100.mL of 0.10M acetate buffer, pH = 4.50? pKa of acetic acid is 4.74

(Abbreviations: acetic acid = HAc, Acetate = Ac)

The first step to solving any buffer problem is to recognize what is the weak acid/base pair that will exist in the buffer solution.  To do this, write out the equilibrium equation for the buffering species.

HAc    +  H2O                H3O+ +    Ac

acid (H+ – attached)                         base (H+ – not attached)

Then identify the weak acid (HA) and base (A) in the reaction.

HA = HAc      A = Ac

Write down the pK associated with this pair. You will be given this information as an actual pK or as a Ka.  If you are given the Ka you must take the –log(Ka) to get the pKa.

pKa for acetic acid / acetate is 4.74

Write out the Henderson-Hasselbach equation using the information given.
pH = pKa + log(A/HA)
4.50 = 4.74 + log(acetate/acetic acid)

Calculate the ratio of A to HA that will exist in your buffer (Solve for A/HA).
A/HA = 10(4.50-4.74) = 10–.24 = 0.575 = 0.58

Next, calculate the total number of millimoles of buffer needed (total acetic acid + acetate or HA + A) by multiplying the volume of the buffer in mL times the molar concentration of the buffer.

100.mL x 0.10 mmols/mL = 10. mmols total buffer.

Determine how to divide the total millimoles into the A and HA forms such that the final ratio will be equal to that calculated (0.58 for this problem). To easily do this, divide the total number of millimoles by 1 + the ratio calculated.

10.millimols / 1.58 = 6.3 mmols

This will be the number of millimoles of HA needed to get the buffer to the proper pH assuming the remainder is in the A form.

HA = 6.3 millimoles = millimoles of acetic acid in 100 mL of buffer

A = 10. millimoles –  6.3 millimoles = 3.7 millimoles of acetate in the 100mL

Check your calculations:

pH = Pka + log(A/HA) = 4.74 + log(3.7 mmols / 6.3 mmols) = 4.51

The small (0.1) error is roundoff error and is acceptable in terms of two significant figures in this calculation.

We must now determine how you will get 6.3 mmols of acetic acid and 3.7 mmols of acetate.  Begin by looking at the starting materials given in the problem. In this problem you are starting with 1.0M acetic acid and 0.50M NaOH. Since you are not given a direct source of the base acetate, you will have to use NaOH to convert some acetic acid into sodium acetate. Remember, one mmol of a strong base like NaOH or KOH will convert an equivalent amount of a weak acid into its conjugate base.  In this case, every mmol of NaOH added will convert an equal number of mmols of HAc into Ac.

Start by adding 10.mL of 1.0M HAc (10.mmols) to a flask.  This will provide all the acetic acid/acetate necessary to give the correct concentration in 100mL of buffer. We then must convert 3.7 mmols of the HAc into Ac.  To do this we must add 7.4mL of 0.50M NaOH (7.4mL x 0.50mmols/mL = 3.7mmols) to the flask.

The flask will now contain a solution whose pH is equal to 4.5. However, the concentration will not be 0.10M. We must add water to the flask to give a final volume of 100.mL.   Mix and our buffer is complete!

The question above is about simple as a buffer problem can be.  However, even more complex problems require one to answer the same basic questions.  Problems usually vary in the following ways:

You are given a base such as an amine and a pKb instead of a pKa. Often the base is in a protonated form, such as ammonium ion or methyl ammonium chloride.  The easiest solution is to write out an acid dissociation reaction and convert the pKb to a pKa by subtracting the pKb from 14. These problems are usually difficult for the student.

You are given a polyprotic acid such as phosphoric acid.  Assumming you start with phosphoric acid the typical phosphate problem will require you to convert all of the phosphoric acid to dihydrogen phosphate (H2PO4–1)and then some of the dihydrogen phosphate into monohydrogen phosphate (HPO4–2) ( most phosphate buffers are around pH=7). Problems using polyprotic acids can also start with intermediates such as H2PO4–1 and HPO4–2 . It is important you write out the equilibrium and determine where you are (starting materials) and where you need to be (A and HA).[wp_campaign_2]

You are given a substance like an amino acid that has both acidic and basic functional groups. Again, writing the equilibrium is crucial to understanding where you are at any given time while solving the problem. In such compounds, pKa’s are normally given and not pKb’s, even when the functional group is a base, like an amine.  Remember, the term HA in the H-H equation refers to the protonated form of the group and A refers to the group as it exists without a proton attached.  For a carboxylic acid group, HA = R-COOH and A = R-COO. For an amino group, HA = R-NH3+ and A = R-NH2 .

Let’s look at another problem:

Starting with solid glycine-HCl and 1.0M NaOH, how would you prepare 375mL of a 0.050M glycine buffer, pH = 9.80? pK1 = 2.34, pK2 = 9.60

Write out the equilibrium:  (glycine-HCl = HOOC–CH2–NH3+ Cl )

HOOC–CH2–NH3+ OOC–CH2–NH3+ OOC–CH2–NH2

pK1=2.34                                    pK2=9.60

Determine HA and A:

For pH = 9.0, you must have significant amounts of both OOC–CH2–NH3+ (HA), and OOC–CH2–NH2 (A) in the solution. This means the acid/base pair is that associated with the pK2 for the substance. There will essentially be no glycine-HCl in the solution at pH=9.0.

Determine the ratio of A/HA:

9.80 = 9.60 + log(A/HA)

A/HA = 100.20 = 1.58/1
Check your ratio:

pH = 9.60 + log(1.58/1) = 9.60 + 0.20 = 9.80   OK

Determine the number of mmols of each species in the buffer:

375mL x 0.050mmols/mL = 18.75mmols total buffer

18.8mmols / 2.58 = 7.27mmols of HA

18.8mmols – 7.27mmols = 11.5mmols of A

Check your ratio:

pH = 9.60 + log(11.8/7.27) = 9.60 + 0.21 = 9.81   OK

Determine the source of A and HA:

All must come from the solid glycine-HCl. All of the glycine-HCl must be converted to OOC–CH2–NH3+ , then some of that must be converted to the fully unprotonated form OOC–CH2–NH2 . This is accomplished by adding NaOH solution.

Do the calculations:

To get 18.8mmols of gly-HCl (MW = 111.5g/mol) you will need to weigh out 2.09 grams (111.5g/mol x 0.0188mol) of solid gly-HCl and place it in a flask.

To convert all 18.8mmols of gly-HCl to the zwitterion form, add 18.8mmols or 18.8mL of 1.0M NaOH to the flask. To convert 11.5mmols of the zwitterions to the unprotonated form (A), add 11.5mL of 1.0M NaOH to the flask.

The ratio of A/HA is now such that the pH is 9.80, simply add water up to a final volume of 375mL and you are done!

For a little exercise, determine the NaCl concentration in the final solution. The NaCl comes from the Na+ from NaOH and the Cl from the gly-HCl

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