Question

Buffer Assignment

pH                                                     concentration (M)                                                     mL

7.8                                                    1.0                                                                                100.00

Eq) Buffer=Stotal/L

Where Stotal=moles (A) + moles (B)

Ptotal=mol (H2PO4-)+mol (HPO42-)

PH=pKa + log (moles B/molesA)

Determine the Mass of Each Component

Recall that buffers are formed from conjugate acid/base pairs. Using the information given about your assigned buffer, determine how much of each component (acid and base) you will need in order to prepare it in the lab. (This will require a system of equations because there are two “unknowns.”)

Note: the conjugate acid in this case is H2PO4−, and the conjugate base is HPO42−.

1. Using equation (3) in your lab manual, solve for the ratio of (moles B/moles A).

2. From the ratio, solve for the moles B in terms of moles A. (Hint: you are not solving for the actual number of moles B at this point, you are only solving for the relationship of B to A.)

3. Using your concentration and volume, solve equation (1) in the lab manual for Stot.  

4. Stot is the total moles of solute components (the sum of your acid and base combined). Using equation (2) in the lab manual, and substituting your answer to question 2 in for “moles B,” solve for the quantity of moles A.

5. Solve for the quantity of moles B by inserting your numerical value for moles A into the relationship found in question 2.

6. Determine how many grams of each component must be measured out in order to make the buffer solution. Note that the conjugate acid comes as a solid of NaH2PO4H2O, and the conjugate base comes as a solid of Na2HPO47H2O.

Determine pH Values of a Buffered System

7. Suppose that you pour half of your buffer solution into a separate beaker and add 1.00 mL of 0.103 M HCl into it. What pH would you expect to measure with a pH probe? Show all work. (Hint: Is HCl an acid or a base? Is it strong or weak? How would it affect your buffer components?)

8. Suppose you add 1.00 mL of 0.098 M NaOH into the other half of your buffer solution. What pH would you expect to measure with a pH probe? Show all work. (Hint: Is NaOH an acid or a base? Is it strong or weak? How would it affect your buffer components?)

Determine pH Values of an Unbuffered System

9. What pH would you expect to measure if you placed 52.5 mL of DI water into a beaker and added 1.00 mL of 0.103 M HCl? Show all work.

10. What pH would you expect to measure if you placed 51.3 mL of DI water into a beaker and added 1.00 mL of 0.098 M NaOH? Show all work.

Answer 1

The ratio of acid and base can be calculated using the Henderson-Hasselbach equation:

Since we know the desired pH (7.8) and can easily find the pKa of the acid/base system (7.2), we can rearrange and solve for the ratio:

The relationship between the number of moles of base and acid is then:

The total number of moles of the buffer can be determined as:

We then have:

If we replace the relaationship between nb and na:

So, we have:

Thus, the number of moles of base can be calculated as: n_b = 0.10 moles - 0.020 moles = 0.080 moles

The mass of acid (NaH₂PO₄.H₂O) will be:

And of base (Na₂HPO_4.7H₂O):

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Acid addition to the buffer: We are adding the following number of moles of acid:

These moles react with the base and convert it to acid, changing their number of moles. We can calculate the new pH using the Henderson-Hasselbach equation (the original number of moles is half what we calculated, since we are using only half of the buffer).

Addition of base to the buffer: We are adding the following number of moles of base:

These moles react with the acid and convert it to base, changing their number of moles. We can calculate the new pH using the Henderson-Hasselbach equation (the original number of moles is half what we calculated, since we are using only half of the buffer).

Addition of water to the acid: We are adding, again, 0.000103 moles of strong acid (fully dissociates) to a final volume of 53.5 mL, making its final concentration:

The pH is then:

Addition of water to the base: We are adding, again, 0.000098 moles of strong base (fully dissociates) to a final volume of 52.3 mL, making its final concentration:

The pH is then: