1. In preparation for doing an experiment, you need to prepare a buffer solution
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Question
1. In preparation for doing an experiment, you need to prepare a buffer solution with a pH of 7.8. A friend in your lab says that she has just the recipe for you, using N-2- hydroxyethylpiperazine-N -2-ethanesulfonic acid (HEPES), a sulfonic acid with a pKa of 7.5. Since you know that the buffer pKa should be close to the desired pH, this sounds good, and you make the solution according to the following recipe, which is intended to make 600 mL of a 0.5 M solution:
Dissolve 0.2 moles of HEPES in 400 mL of water.
Add 0.1 moles of Na-HEPES (the sodium salt of the conjugate base) and dissolve. Add water to a total volume of 600 mL.
(a) (4 pts.) Before adding water to the final volume, you decide to check the pH. It’s a good thing you did, because it looks like your friend gave you the wrong recipe! Calculate the expected pH for the solution, made according to the recipe.
(b) (3 pts.) Now, you want to change the pH to the desired 7.8, but you don’t want to add any more HEPES or Na-HEPES. You have on hand stock solutions of 1 M HCl and 1 M NaOH. Which of these solutions should you add to change the pH to 7.8? Explain what will happen to the different components of the solution when you add the HCl or NaOH.
(c) (5 pts) How many mL of the 1 M HCl or NaOH solution should you add?
Explanation / Answer
The pH of the buffer is given by the Henderson-Hasslebach equation as
pH = pKa + log [Na-HEPES]/[HEPES]
a) We added 0.2 mole HEPES and 0.1 mole of Na-HEPES in 400 mL water and dissolved. The pH of the buffer thus prepared is given as
pH = 7.50 + log [(0.1 mole)/400 mL]/[(0.2 mole)/400 mL]
= 7.50 + log (0.1/0.2) = 7.50 + log (0.5)
= 7.50 + (-0.30) = 7.50 – 0.30 = 7.20 (ans).
b) Use the Henderson-Hasslebach equation to calculate the ratio of Na-HEPES and HEPES in the desired buffer.
7.80 = 7.50 + log [Na-HEPES]/[HEPES]
====> 0.30 = log [Na-HEPES]/[HEPES]
====> [Na-HEPES]/[HEPES] = antilog(0.30) = 1.995 2.0
====> [Na-HEPES] = 2.0*[HEPES] ……(1)
Again, the total buffer concentration is 0.5 M; hence, we must have,
[Na-HEPES] + [HEPES] = 0.5 M
====> 2*[HEPES] + [HEPES] = 0.5 M
====> 3*[HEPES] = 0.5 M
====> [HEPES] = (0.5 M)/3 = 0.1666 M 0.17 M.
Therefore, [Na-HEPES] = 2*[HEPES] = 2*(0.17 M) = 0.34 M.
We have 600 mL buffer; hence,
moles Na-HEPES = (600 mL)*(1 L/1000 mL)*(0.34 M) = 0.204 mole.
moles HEPES = (600 mL)*(1 L/1000 mL)*(0.17 M) = 0.102 mole.
The desired buffer must contain 0.102 mole 0.10 mole HEPES and 0.204 mole 0.20 mole Na-HEPES. However, we added 0.2 mole HEPES and 0.1 mole Na-HEPES. Therefore, we must increase the number of moles of Na-HEPES in the buffer. This can be done by adding 1 M NaOH to the buffer solution prepared in (a) above.
c) The neutralization reaction between HEPES and NaOH is given as
HEPES + NaOH ---------> Na-HEPES + H2O
As per the stoichiometry of the reaction,
1 mole HEPES = 1 mole NaOH = 1 mole Na-HEPES.
We must have (0.2 – 0.1) mole = 0.1 mole Na-HEPES produced by the neutralization reaction. Therefore, we must add 0.1 mole NaOH to the buffer prepared in (a).
Volume of NaOH added = (0.1 mole)/(1 M) = 0.1 L = (0.1 L)*(1000 mL/1 L) = 100 mL (ans).
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