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?

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...n 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 ...