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# Chemistry

### Determining the pH of an acid and base in various situations by working with concentrations. A multi-step question.

Part 1: What is the pH of a solution of 1M HCO3 and 1M CO3?

Part 2: What is the pH if it was mixed with an equal volume of water?

Part 3: What is the pH if it was mixed with an equal volume of 0.001 M HCl?

Part 4: What is the pH if it was mixed with an equal volume of 0.001 M NaOH?

### Solution

PART ONE:

"What is the pH of a solution of 1M HCO3 and 1M co3?"

Step 1:

To calculate the pH of the solution, you will need to understand the Henderson-Hasselbach equation, and how buffers work, in general. You have probably learned that a "buffer" is a solution that is composed of a mixture of weak acid and weak base. The pH of the solution is determined by (a) the ratio of weak acid and weak base in the solution, and (b) the pKa of the weak acid. The relation of these factors to one another is defined by the "Henderson-Hasselbach equation" (for convenience, I will call this the "HH equation" from now on.)

The HH equation is useful in calculations involving situations such as (i) titrations of weak acids or bases with strong acids or bases, (ii) preparation of buffers in the laboratory, (iii) calculating the theoretical pH of solutions of weak acids/bases, (iv) etc. The equation is as follows:

pH = pKa + log ([A-]/[HA]), where

pH = the pH of the solution
pKa = -log (Ka)
[A-] = the concentration of weak base in the
solution
[HA]=the concentration of weak acid in the
solution

Step 3: Identify which species is the "weak acid", and which is the "weak base". Write a chemical equation to describe the relationship between the two. (Writing the chemical equation helps out with answering PART THREE, below, and also helps you understand which species is acid vs. base.)

In the problem you have given, the species HCO3(1-) is the "weak acid", and
CO3(2-) is the "weak base".

The chemical equation for the equilibrium between weak acid and weak base would be:

HCO3(1-) * H(1+) + CO3(2-)

Step 4: Look up the pKa of your weak acid.

pKa values can be found in textbooks. Many professors will provide pKa (or Ka) values/tables on exams. However, some professors will expect you to memorize important pKa values --- check with your instructor!

In "Biochemical Calculations" (2nd edition, I.H.Segel), the pKa value for the above reaction is 10.25.

Step 5: Use the pKa value, and the HH equation, to calculate the pH:

pH = pKa + log ([A-]/[HA])
pH = 10.25 + log (1M/1M)
pH = 10.25 + log (1)
pH = 10.25 + 0
pH = 10.25

SHORTCUT: Going through the above exercise illustrates the method for calculating the pH of a buffered solution. Now, here is a convenient shortcut: For any buffer, when the concentration of weak acid is equal to the concentration of weak base, pH = pKa. Try the calculation for another buffer to prove it to yourself.

PART TWO:
"What is the pH if it was mixed with an equal volume of water?"

If the solution described above was mixed with an equal volume of water, what would the new concentration of HCO3(1-) be? Of CO3(2-) be? The quick answer is that adding an equal volume of water will dilute each concentration by a factor of 2, so the new[HCO3](1-) would be 0.5M, and the new[CO3](2-) would be 0.5 M.

If you have trouble visualizing the dilution using the quick method, you can use the following equation to calculate the new concentrations: C1xV1=C2xV2, where C1=old concentration, V1=old volume, C2=new concentration, V2=new volume. For example, if the old solution had a volume of 1 L: (1M)x(1L)=(C2)x(2L).......C2=1ML/2L=0.5M

Getting to the actual question....what is the new pH?

Once again, use the HH equation. You know the pKa from part one of this question, and you know the new concentrations, so you can calculate the new
pH:

pH = 10.25 + log (0.5M/0.5M) = 10.25 + log(1) = 10.25

The pH didn't change! Why not? If you think about this, it makes sense...when water was added, the concentration of the weak acid and weak base changed by the same amount. The ratio of weak acid to weak base was 1 when you started, and 1 after the addition of water.......since the ratio is still one, pH still equals pKa. So, the pH of a buffer doesn't change when it is diluted. (The buffering CAPACITY does, but that's a different question.)

PART THREE:

"What is the pH if it was mixed with an equal volume of 0.001 M HCl?" (Presumably, we are talking about the original "1M + 1M" buffer here (ie. before dilution with water). If not, adjust the concentrations
appropriately.)

First, let's rewrite the chemical equation for the equilibrium between weak acid and weak base.

HCO3(1-) * CO3(2-) + H(1+)

HCl is a strong acid. Will a strong acid react with the weak acid or weak base? It will react with the weak base. Write the chemical equation for the reaction with HCl:

CO3(2-) + H(1+) * HCO3(1-)

(I turned the equation around, so that it is easier to think about "adding acid".)

We'll come back to the equation in a moment. In the mean time, let's calculate the amount of HCO3(1-) and CO3(2-) in the buffer BEFORE HCl is added. It's easier to think of in terms of "amount", rather than "concentration", so let's pretend we have 1L of buffer (any volume could be used - it is arbitrary).

BEFORE HCl is added, amount of CO3(2-) = 1 mole/L x 1 L = 1 mole BEFORE HCl is added, amount of HCO3(1-) = 1 mole/L x 1L = 1 mole

Next, calculate how much HCl is added. Since an equal volume is added, the amount of HCl added = 1L x 0.001 mole/L = 0.001 mole.

Now, make a table under the chemical equation for the reaction with HCl. The table will help us sort out what changes are happening when HCl is added. Eventually, you will be able to understand what is happening well enough to do this without writing it down, but for now, the table is a good learning tool. There are 5 lines in the table - (a) the "before" case, where we enter the amount of each species before any acid is added; (b) the "added" case, where we enter the amount of acid that is added; (c) the "change" case, where we figure out what changes (or "what reacts with what") are caused by the addition in (b); and (d) the "final" case, where we calculate the amount of each species after the reaction with HCl is complete. (To calculate the amounts in (d), just add up the numbers in each column.) The last line, (e), is the FINAL amount of each species expressed as a concentration, instead of as the number of moles. In this case, we are thinking about 1 L of solution, so FINAL (conc) = FINAL (moles) divided by 1L. Here we go:

CO3(2-) + H(1+) * HCO3(1-)
(a) BEFORE:1 mole + 0 * 1 mole
(b) ADDED : 0 + 0.001 mole * 0
(c) CHANGE: -0.001 mole + (-0.001 mole) + 0.001 mole (b/c it reacted with H+)
----------------------------------------------------------------------------
-----------------
(d) FINAL:0.999 mole 0 1.001 mole
(e) FINAL (conc) 0.999 M 0 1.001 M

Now....What is the new pH? Use the HH equation. The concentrations of weak acid and weak base are obtained from line (e), above.

pH = pKa + log ([CO3(2-)]/[HCO3(1-)])
pH = 10.25 + log(0.999M/1.001M)
pH = 10.25 + log(0.998)
pH = 10.25

The pH did not change! Why not? The buffering capacity is very strong compared to the amount of HCl that is added.

PART FOUR:

Answer this question in exactly the same way as part three, above. NaOH is a strong base, so it will react with the weak acid (HCO3(1-)).