You are asked to prepare 500. mL of a 0.200 M acetate buffer at pH 4.90 using on

Question

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.

1. How many grams of acetic acid will you need to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.

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.)

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 1. How many grams of acetic acid will you need to prepare the 500 mL buffer? Note that the given concentration 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? O 2 20

Explanation / Answer

a) Molarity= mol ste/ V(L)

mol= M*V= 0,2M*0,5L= 0,1mol

1mol CH3COOH ---------60,05g

0,1mol ----------------------X= 6,005g

b) We have the following equilibrium

CH3COOH ======= CH3COO- + H+

At the equilibrium you can have

Ka= [CH3COO-][H+]/[CH3COOH]

[CH3COO-]=[H+]

[CH3COO-]=( Ka*[CH3COOH])1/2

[CH3COO-]= 0,002M

To prepare the buffer you use the following equation

pH = pKa + log [A-]/[HA]

4,90 = 4,76 + log [A-]/[HA]

[A-]/[HA] =10 4,90- 4,76

[A-]/[HA] = 1,38

So the ratio between the conjugated base and the acid must be 1,38 to get a pH of 4,90.

Now, in the relation [A-]/[HA] = 1,38 we can calculate [A-]

[A-]= [HA]*1,38 = 0,2M* 1,38= 0,28M

We need to have a concentration of H+ or CH3COO- of 0,28M to get the pH=4,90 but at the equilibrium we only have 0,002M of this 2 species, so we add as much NaOH to consume the H+ to get equilibrium change to the product.

So 0,28M-0,002M= 0,278M, this 0,278M is the concentration of OH- that we will need to add to make the equilibrium change.

So M= mol/V

mol= 0,278M*0,5L=0,139mol

So this are the mols of NaOH that you need to prepare the solution of 500ml at pH of 4,90.

Now we can calculate the volume we need:

M=mol/V ----------V= mol/M= 0,139/3,00M= 0,046L

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