The K_aof acetic acid is 1.76 × 10^-⁵. The pH of a buffer prepared by combining 50.0 mL of 1.00 M potassium acetate and 50.0 mL of 1.00 M acetic acid is _ . A) 0.85 B) 2.38 C) 4.77 D) 1.70 E) 3.40

Question

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To find the pH of a buffer, we need to use the Henderson-Hasselbalch equation:

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

where pKa is the dissociation constant of the weak acid, [A-] is the concentration of the conjugate base (in this case, potassium acetate), and [HA] is the concentration of the weak acid (acetic acid).

From the given information, we can find the pKa of acetic acid:

pKa = -log(Ka) = -log(1.76 × 10^-5) = 4.76

Next, we need to find the concentrations of [A-] and [HA] in the buffer solution. We know that we combined 50.0 mL of 1.00 M potassium acetate, so the moles of potassium acetate are:

n(A-) = 0.0500 L x 1.00 mol/L = 0.0500 mol

Assuming that potassium acetate dissociates completely to K+ and A-, the concentration of A- is also 0.0500 M.

For acetic acid, we need to take into account that it does not dissociate completely in solution. We can use the given Ka and the initial concentration of acetic acid to set up an ICE table:

Ka = [H+][A-]/[HA]
1.76 × 10^-5 = (x)(0.0500)/(0.0500 - x)

where x is the concentration of H+ and (0.0500 - x) is the concentration of undissociated acetic acid.

Solving for x, we get:

x = 8.55 × 10^-5

This concentration of H+ corresponds to a pH of:

pH = -log(H+) = -log(8.55 × 10^-5) = 4.07

To check our assumption that the concentration of undissociated acetic acid is negligible compared to the initial concentration of 1.00 M, we can calculate the percent dissociation:

% dissociation = ([H+] / initial [HA]) x 100
% dissociation = (8.55 × 10^-5 / 1.00) x 100
% dissociation = 0.00855 x 100 = 0.855%

This is a very low percent dissociation, so our assumption is valid.

Now we can use the Henderson-Hasselbalch equation to find the pH:

pH = pKa + log([A-]/[HA])
pH = 4.76 + log(0.0500/0.00855)
pH = 4.76 + 1.31
pH = 6.07

However, we need to remember that we calculated the pH of the acetic acid solution, not the buffer solution. To find the pH of the buffer, we need to take into account the dilution of the acetic acid by the potassium acetate:

moles of acetic acid = initial moles / dilution factor
moles of acetic acid = 0.0500 mol / 2
moles of acetic acid = 0.0250 mol

moles of potassium acetate = initial moles / dilution factor
moles of potassium acetate = 0.0500 mol / 2
moles of potassium acetate = 0.0250 mol

total volume of buffer solution = 50.0 mL + 50.0 mL = 0.1000 L

concentration of acetic acid in buffer = moles / volume
concentration of acetic acid in buffer = 0.0250 mol / 0.1000 L
concentration of acetic acid in buffer = 0.250 M

concentration of potassium acetate in buffer = moles / volume
concentration of potassium acetate in buffer = 0.0250 mol / 0.1000 L
concentration of potassium acetate in buffer = 0.250 M

Now we can use the Henderson-Hasselbalch equation to find the pH of the buffer:

pH = pKa + log([A-]/[HA])
pH = 4.76 + log(0.250/0.250)
pH = 4.76 + 0
pH = 4.76

Therefore, the answer is C) 4.77.

The Ka of Acetic Acid Is 1

The K_aof Acetic Acid Is 1