A solution buffered at a pH of 5.00 is needed in an experiment. Can we use acetic acid and sodium acetate to make it? If so, how many moles of NaC2H3O2 must be added to a 1.0L of a solution that contains 1.0 mol HC2H3O2 to prepare the buffer. The Ka of acetic acid is 1.8x10^-5.

HC2H3O2 + H2O = H3O+ + C2H3O2

I got an answer of 1.82 moles C2H3O2

That's probably right. If I use 4.76 for pKa for acetic acid and plug in 1.82 for mols base and 1 for mols acid, I get pH 5.02. Using pKa 4.74 gives 5.00.

To determine if acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2) can be used to make a solution buffered at a pH of 5.00, we need to consider the pKa value of acetic acid.

The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa and the ratio of the concentrations of the conjugate acid and base:

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

For a pH of 5.00, we need to calculate the ratio [A-]/[HA] using the pKa value and solve for [A-].
Given Ka = 1.8x10^-5, the pKa value is:

pKa = -log(Ka)
pKa = -log(1.8x10^-5)
pKa ≈ 4.74

Using the Henderson-Hasselbalch equation, substituting the given pH and pKa values:

5.00 = 4.74 + log([A-]/[HA])
0.26 = log([A-]/[HA])

To make a buffer, we want to start with 1.0 mol of acetic acid (HC2H3O2) and determine how many moles of sodium acetate (NaC2H3O2) are needed. The moles of [A-] (sodium acetate) will be equal to the moles of [HA] (acetic acid) according to the balanced equation:

HC2H3O2 + NaOH → H2O + NaC2H3O2

Since we start with 1.0 mol of acetic acid, we need to find the number of moles of sodium acetate needed. From the Henderson-Hasselbalch equation, the ratio [A-]/[HA] is given as:

[A-]/[HA] = 10^(0.26)
[A-]/[HA] ≈ 1.82

So, to prepare the buffer, you would need to add approximately 1.82 moles of sodium acetate to a 1.0 L solution containing 1.0 mol of acetic acid.

Yes, you can use acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2) to make a solution buffered at a pH of 5.00.

To calculate the amount of NaC2H3O2 needed to prepare the buffer solution, you need to consider the Henderson-Hasselbalch equation:

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

In this equation, pH is the desired pH of the buffer solution, pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base (C2H3O2- in this case), and [HA] is the concentration of the acid (HC2H3O2).

In this case, the pKa of acetic acid is given as 1.8x10^-5, and the desired pH is 5.00. By substituting these values into the Henderson-Hasselbalch equation, we can solve for [A-]/[HA].

5.00 = -log(1.8x10^-5) + log([A-]/[HA])

Let's rearrange the equation:

log([A-]/[HA]) = 5.00 + log(1.8x10^-5)

Now, let's take the antilog of both sides to obtain [A-]/[HA]:

[A-]/[HA] = 10^(5.00 + log(1.8x10^-5))

[A-]/[HA] = 10^5.00 * 10^log(1.8x10^-5)

[A-]/[HA] = 10^5.00 * 1.8x10^-5

[A-]/[HA] = 17.8

Since we already have 1.0 mol of HC2H3O2, we can assume the concentration is also 1.0 M. So, we can set [HA] = 1.0 M.

Now, to find the concentration of [A-], we multiply the ratio by the concentration of [HA]:

[A-] = [A-]/[HA] * [HA]

[A-] = 17.8 * 1.0 M

[A-] = 17.8 M

Finally, to calculate the moles of NaC2H3O2 required, we multiply the concentration ([A-]) by the volume (1.0 L):

moles of NaC2H3O2 = [A-] * volume

moles of NaC2H3O2 = 17.8 M * 1.0 L

moles of NaC2H3O2 = 17.8 moles

Therefore, you would need to add 17.8 moles of NaC2H3O2 to a 1.0L solution containing 1.0 mol of HC2H3O2 in order to prepare the buffer at a pH of 5.00.