You have a #"250.0-mL"# sample of #"1.00-M"# acetic acid. Assuming no volume change, how much #"NaOH"# must be added to make the best buffer? Calculate the #"pH"# of the best buffer.

In this context, the "best buffer" is a buffer that contains equal concentrations of acetic acid, the weak acid, and of acetate anions, the conjugate base.

Here

Your goal here will be to figure out how much sodium hydroxide must be added to your initial solution in order to get

Your initial solution contains

Now, after the reaction is complete, you want the resulting solution to contain equal numbers of moles of acetic acid and of acetate anions.

To convert this to grams, use the molar mass of sodium hydroxide.

The answer is rounded to three sig figs, the number of sig figs you have for the concentration of the initial solution.

After the reaction is complete, the buffer will contain--keep in mind that we assume that the volume of the buffer will not change upon the addition of the sodium hydroxide

and

which is equal to

The answer is rounded to two decimal places because you have two sig figs for the acid dissociation constant.

To create the best buffer, we need to add equal moles of acetic acid (CH3COOH) and sodium acetate (CH3COONa). Since acetic acid is a weak acid, we can assume it will only partially dissociate. The dissociation equation is:

[CH_3COOH \rightleftharpoons CH_3COO^- + H^+]

Given that the initial concentration of acetic acid is (1.00 , M) and we have (250.0 , mL), we have (0.25 , mol) of acetic acid.

To find the amount of NaOH needed, we first calculate the moles of acetic acid that dissociate. Since acetic acid is a weak acid, it dissociates partially according to the equilibrium expression:

[K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}]

Given that the Ka for acetic acid is (1.8 \times 10^{-5}), we can use the ICE table (Initial, Change, Equilibrium) to determine the concentration of (H^+) ions.

[K_a = \frac{x^2}{0.25-x}]

We can approximate (0.25 - x \approx 0.25), so:

[1.8 \times 10^{-5} = \frac{x^2}{0.25}]

Solving for (x), we find (x \approx 7.54 \times 10^{-4}) M, which represents the concentration of (H^+) ions.

Since acetic acid and sodium acetate will react in a 1:1 ratio, we need the same concentration of sodium hydroxide (NaOH) to react with the (H^+) ions. Therefore, we need (7.54 \times 10^{-4}) M NaOH.

To find the volume of NaOH needed, we use the formula:

[M_1V_1 = M_2V_2]

Substituting the given values, we have:

[(1.00 , M)(V_1) = (7.54 \times 10^{-4} , M)(0.250 , L)]

Solving for (V_1), we find (V_1 \approx 1.89 \times 10^{-4} , L) or (0.189 , mL) of NaOH.

The pH of the buffer can be calculated using the Henderson-Hasselbalch equation:

[pH = pKa + \log\left(\frac{[A^-]}{[HA]}\right)]

Where (pKa) is the negative logarithm of the acid dissociation constant ((Ka)), ([A^-]) is the concentration of the conjugate base (acetate ions), and ([HA]) is the concentration of the weak acid (acetic acid).

Given that (pKa = -\log(Ka)), and (Ka = 1.8 \times 10^{-5}), we find (pKa \approx 4.74).

Plugging in the values, we get:

[pH = 4.74 + \log\left(\frac{0.25}{0.25}\right)]

[pH = 4.74 + \log(1)]

[pH = 4.74]