help with question number 2 Name: WEEK 2 Part F: Pre-lab Calculations-Week 2 Dat
ID: 1046959 • Letter: H
Question
help with question number 2
Name: WEEK 2 Part F: Pre-lab Calculations-Week 2 Date Score: 25 Prepare 50 ml of an acetic acid/sodium acetate buffer that has a pH of 5 and a 0.1 M concentration of the acid. Devise a plan for making this buffer from acetic acid solution and solid sodium acetate. The goal is to calculate the mass of sodium acetate you need to weigh out and dissolve in the acetic acid that would give you a solution with your target pH 1. Consider the Henderson-Hasselbalch equation log HA Using the pk, acetic acid from Appendix A in the lab handout, the target pH of S, and the 0.1 M concentration of acetic acid (HA in the HH-equation), calculate the concentration of sodium acetate (A in the HH-equation) P Pka + log (4 O IM From the concentration of sodium acetate, first calculate the moles of sodium acetate (remember that you know the total volume of your solution), and then calculate the number of grams you need to prepare your buffered solution 0100885 mo) roar o ?26 concentration of your unknown acid. Devise a plan for making this buffer from your unknown acid and 0.I M NaOH. The goal is to calculate the initial mass of unknown acid you need to weigh out and also the volume of 0.1 M NaOH needed to give the buffered solution with the target pH. 2. Prepare 50 mL of your unknown acid/conjugate base buffer that has a pH of 5 and a 0.1 M This is somewhat trickier mathematically since the unknown acid is a solid and the conjugate base, A, of the acid will be created in solution using sodium hydroxide to convert HA to A. The amount of unknown acid must be calculated as the total of what is needed to be 0.1 M at equilibrium and what is needed to create enough conjugate base to buffer the 50 mL of solution at a pH of 5 en-mcneil 90000Explanation / Answer
2) The pH of the buffer is 5. Again, a good choice for the buffer is acetic acid, HA (pKa = 4.75) and sodium acetate, NaA. Use the Henderson-Hasslebach equation to calculate the molarity of NaA in the buffer (the molarity of HA in the buffer is 0.1 M, given).
pH = pKa + log [NaA]/[HA]
====> 5 = 4.75 + log [NaA]/[HA]
====> log [NaA]/[HA] = 5 – 4.75 = 0.25
====> [NaA]/[HA] = antilog (0.25) = 1.778
====> [NaA] = 1.778*[HA]
Again,
[NaA] + [HA] = 0.1 M
====> 1.778*[HA] + [HA] = 0.1 M
====> 2.778*[HA] = 0.1 M
====> [HA] = (0.1 M)/(2.778) = 0.036 M
Therefore,
[NaA] = 1.778*[HA] = 1.778*(0.036 M) = 0.064 M.
We have 50 mL of the buffer; hence, millimole(s) of NaA in the buffer = (50 mL)*(0.064 M) = 3.2 mmole.
NaA is produced by the action of 0.1 M NaOH on HA as below.
HA (aq) + NaOH (aq) -------> NaA (aq) + H2O (l)
As per the stoichiometric equation,
1 mole HA = 1 mole NaOH = 1 mole NaA.
Therefore,
3.2 mmole NaA = 3.2 mmole NaOH = 3.2 mmole HA.
Volume of 0.1 M NaOH required to produce 3.2 mmole NaA = (millimoles of NaOH)/(molarity of NaOH) = (3.2 mmole)/(0.1 M) = 32 mL.
MilliMole(s) of HA present in the buffer = (50 mL)*(0.036 M) = 1.8 mmole.
Total millimoles of HA taken to prepare the buffer = (3.2 + 1.8) mole = 5.0 mmole.
Molar mass of acetic acid (C2H4O2) = (2*12.011 + 4*1.008 + 2*15.999) g/mol = 60.052 g/mol.
Mass of acetic acid needed to be weighed out = (5.0 mmole)(1 mole/1000 mmole)*(60.052 g/mol) = 0.30026 g ? 0.300 g.
Therefore, in order to prepare 50 mL of the buffer, weigh out 0.300 g solid acetic acid, add 32 mL of 0.1 M NaOH and make upto the mark with DI water.
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