You are asked to prepare 500 ml 0.250 m acetate buffer at ph 5.00 using only pure acetic acid ( mw=60.05 g/mol, pka=4.76), 3.00 m naoh, and water. answer the questions regarding the preparation of the buffer.

1. how many grams of acetic acid will be needed to prepare the 500 ml buffer?
note that the given concentration of acetate refers to the concentration of all acetate species in solution.

7.506 g

Explanation:

In order to answer the first question we need to keep in mind the requested concentration of acetate: 0.250 M. With the concentration and the volume we can calculate the required moles and then the required grams:

0.250 M * 0.500 L = 0.125 moles acetate

0.125 moles * 60.05 g/mol = 7.506 g

So 7.506 g of acetic acid will be needed to prepare the buffer.

Explanation:

In order to do this, we need to use the Henderson - Hasselbach equation which is the following:

pH = pKa + log[A^-]/[HA] (1)

This equation is used so we can calculate the ratio of A-/HA and determine later the concentration of HA needed, in this case, the acetic acid.

So, let's use (1) to solve for ratio:

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

[A^-]/[HA] = 10^(pH - pKa) (2)

Now, let's replace the given data of pH and pKa and let's calculate the ratio:

ratio = 10^(5-4.76)

ratio = 1.7378

Now, let's calculate the moles of the buffer solution:

moles buffer = moles A + moles HA = 0.25 mol/L * 0.5 L = 0.125 moles (3)

With the obtained ratio, we can have a ratio of moles too:

[A-]/[HA] = moles A/V / moles HA/V ---> from here, V can be cancelled out:

moles A / moles HA = 1.7378

solving for moles A:

moles A- = 1.7378molesHA

now, we can replace this value in (3) and solve for the moles of HA:

0.125 = moles A- + molesHA

0.125 = 1.7378moles HA + moles HA

0.125 = 2.7378molesHA

molesHA = 0.125/2.7378

molesHA = 0.0457 moles

These are the moles of the acetic acid needed to prepare the solution, so, sll we need to do is use the molar mass of the acetic acid and get the mass needed:

m = 0.0457 * 60.05

mass of acetic acid = 2.7443 g

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.