You wish prepare 50.0 mL of an acetate buffer with a concentration of 200 mM and
ID: 530362 • Letter: Y
Question
You wish prepare 50.0 mL of an acetate buffer with a concentration of 200 mM and a pH of to acetate (molecular weight 60.05 g/mol) and 1.0 5.2. You have solid sodium M acetic acid on hand. The pKa of acetic acid is 4.75. Show your work Box your final answer. Include units. a. What will you mix (and in what quantities) to prepare this buffer? b. Despite the careful planning you did in part la, your lab has run out of sodium acetate and the Harvard post doc has broken the electrode on the pH meter. Again. Why isn't he more careful with the stir bar?? Luckily, you have a can-do spirit, the 1.0 M acetic acid, 3.0 M sodium hydroxide, 3.0 M hydrochloric acid, and a thorough understanding of biochemistry. What will you mix (and in what quantities) to prepare this buffer?Explanation / Answer
2a) The pH of acetic acid/sodium acetate buffer is given by the Henderson-Hasslebach equation as
pH = pKa + log [NaOAc]/[HOAc] (HOAc = acetic acid; NaOAc = sodium acetate)
Plug in values and obtain
5.20 = 4.75 + log [NaOAc]/[HOAc]
===> 0.45 = log [NaOAc]/[HOAc]
===> [NaOAc]/[HOAc] = antilog (0.45) = 2.818 2.82
===> [NaOAc] = 2.82*[HOAc] …..(1)
Again, we have [NaOAc] + [HoAc] = 200 mM = 200*10-3 M
===> 2.82*[HOAc] + [HOAc] = 0.2 M
===> 3.82*[HOAc] = 0.2 M
===> [HOAc] = 0.2/3.82 M = 0.052 M
Therefore, [NaOAc] = 2.82*0.052 M = 0.1466 M 0.147 M
Now that we have deduced the required concentrations of HOAc and NaOAc to prepare the buffer, we can find out the amount of solid NaOAc and stock HOAc required.
Volume of the solution = 50 mL = (50 mL)*(1 L/1000 mL) = 0.05 L.
Moles of NaOAc required = (0.05 L)*(0.147 mol/L) = 0.00735 mole.
Mass of NaOAc required = (0.00735 mole)*(60.05 g/mol) = 0.44136 g 0.441 g (ans).
Use the dilution equation to find out the volume of stock HOAc required.
V1*(1.0 M) = (50 mL)*(0.052 M)
===> V1 = (50*0.052)/(1.0) mL = 2.6 mL.
We will need to take 2.6 mL of stock 1.0 M acetic acid solution, add 0.441 g solid NaOAc, dissolve the solid and make up the volume to the 50 mL mark with water (ans).
b) We have 1.0 M acetic acid. We need to prepare sodium acetate in the solution by the action of NaOH on acetic acid. The equation taking place is
NaOH + HOAc ------> NaOAc + H2O
As per the stoichiometric equation,
1 mole NaOAc = 1 mole HOAc = 1 mole NaOAc.
At equilibrium, we must have 0.00735 mole NaOAc to successfully form the buffer. We must find out the moles of HOAc retained at equilibrium = (0.05 L)*(0.052 mol/L) = 0.0026 mole.
HOAc is neutralized by NaOH in a 1:1 ratio; the moles of HOAc neutralized = moles of NaOAc formed = 0.00735.
Therefore, total moles of HOAc that we started with = (moles of HOAc neutralized) + (moles of HOAc retained at equilibrium) = (0.00735 + 0.0026) mole = 0.00995 mole.
Concentration of HOAc present initially = (total number of moles of HOAc)/(volume of buffer) = (0.00995 mole)/(0.05 L) = 0.199 mol/L 0.200 M.
Use the dilution equation to find out the volume of stock HOAc required.
V2*(1.0 M) = (50 mL)*(0.2 M)
===> V2 = (50*0.2)/(1.0) mL = 10 mL
Moles of 3.0 M NaOH required for neutralization of 0.00735 mole HOAc = moles of HOAc neutralized = 0.00735 mole.
Volume of 3.0 M NaOH required = (0.00735 mole)/(3.0 mol/L) = 0.00245 L = (0.00245 L)*(1000 mL/1 L) = 2.45 mL.
Therefore, to prepare the buffer, we must take 10.0 mL of stock 1.0 M acetic acid, add 2.45 mL of 3.0 M NaOH solution and dilute upto the 50 mL mark with water (ans).
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