Mapoob Sapling Learning You are asked to prepare 500. mL of a 0.200 M acetate bu
ID: 1045725 • Letter: M
Question
Mapoob Sapling Learning You are asked to prepare 500. mL of a 0.200 M acetate buffer at pH 4.90 using only pure acetic acid (MW-60.05 g/mol, pKa-4.76), 3.00 M NaOH, and water. Answer the following questions regarding the preparation of the buffer. of acetate refers to the concentration of all acetate species in solution. Number 2. What volume of 3.00 M NaOH must you add to the acetic acid to achieve a buffer with a pH of 4.90 at a final volume of 500 mL? (Ignore activity coefficients.) Number mL 3. Pour the beaker contents into a 500 mL volumetric flask. Rinse the beaker several times and add the rinses to the flask. Swirl the flask to mix the solution. Add water to the mark to make the volume 500 mL. Stopper the flask and invert how many times to ensure complete mixing? 2 20Explanation / Answer
Denote acetic acid as AcOH and sodium acetate as NaOAc.
Use the Henderson-Hasslebach equation to determine the ration of NaOAc and HOAc at pH = 4.90.
pH = pKa + log [NaOAc]/[HOAc]
====> 4.90 = 4.76 + log [NaOAc]/[HOAc]
====> log [NaOAc]/[HOAc] = 4.90 – 4.76 = 0.14
====> [NaOAc]/[HOAc] = antilog (0.14) = 1.380
====> [NaOAc] = 1.380*[HOAc] …… (1)
The total acetate concentration is given as 0.200 M; therefore,
[NaOAc] + [HOAc] = 0.200 M
=====> 1.380*[HOAc] + [HOAc] = 0.200 M
=====> 2.380*[HOAc] = 0.200 M.
=====> [HOAc] = (0.200 M)/(2.380) = 0.084 M.
Therefore, [NaOAc] = 1.380*[HOAc] = 1.380*(0.084 M) = 0.11592 M.
We prepare 500 mL of the buffer solution. NaOAc is obtained directly; rather is obtained by adding NaOH to HOAc to produce NaOAc as below.
NaOH (aq) + HOAc (aq) -------> NaOAc (aq) + H2O (l)
As per the stoichiometric equation,
1 mole NaOH = 1 mole HOAc = 1 mole NaOAc.
1) Determine the mole(s) of NaOAc in the buffer.
Mole(s) of NaOAc in the buffer = (500 mL)*(1 L/1000 mL)*(0.11592 M) = 0.05796 mole = moles of HOAc neutralized.
Moles of HOAc in the buffer present = moles of HOAc retained + moles of HOAc neutralized = (500 mL)*(1 L/1000 mL)*(0.084 M) + 0.05796 mole = 0.042 mole + 0.05796 mole = 0.09996 mole.
Mass of HOAc required to prepare the buffer = (0.09996 mole)*(60.05 g/mol) = 6.002598 g ? 6.002 g.
2) Moles of NaOH = moles of HOAc neutralized = 0.05796 mole.
Volume of 3.00 M NaOH required = (0.05796 mole)/(3.00 M) = 0.01932 L = (0.01932 L)*(1000 mL/1 L) = 19.32 mL (ans).
3) The flask must be inverted 20 times to ensure complete mixing (since, the more times of invert, the better are the chances of mixing).