A pH 4.88 buffer was prepared by dissolving 0.10 mol of benzoic acid (K_a = 6.3 × 10^-5)and 0.50 mol of sodium benzoate in sufficient pure water to form a 1.00 L solution.To a 70.0 mL aliquot of this solution was added 2.00 mL of 2.00 M aqueous HI solution.What was the pH of the new 72.0 mL solution? A)2.84 B)3.16 C)3.36 D)4.65 E)4.90

Question

Quizplus · Accepted Answer

First, we need to find the initial pH of the buffer solution.

pKa of benzoic acid = 4.88

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

where [A-] is the concentration of sodium benzoate and [HA] is the concentration of benzoic acid.

Substituting the given values,

4.88 = 4.88 + log (0.50/0.10)

log (0.50/0.10) = 0

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

This means that the buffer is at equal concentrations of its conjugate acid-base pair and the pH is equal to the pKa (4.88 in this case).

Next, we need to find the new concentrations of the buffer components after addition of HI.

HI reacts with sodium benzoate to form benzoic acid and sodium iodide:

C6H5CO2Na + HI -> C6H5COOH + NaI

We can use the balanced chemical equation to find the moles of sodium benzoate that react with the added HI:

2.00 mL of 2.00 M HI solution = 0.00400 mol of HI

From the balanced equation, we can see that 1 mole of HI reacts with 1 mole of sodium benzoate. Since we added only a small amount of HI (2.00 mL of 2.00 M), we can assume that the amount of sodium benzoate that reacted is small relative to the initial amount present in the buffer solution.

So the new concentration of sodium benzoate can be approximated as:

New [A-] = 0.50 mol/L - (0.00400 mol/L) = 0.496 mol/L

The new concentration of benzoic acid can be approximated as:

New [HA] = 0.10 mol/L + (0.00400 mol/L) = 0.104 mol/L

Now, we can calculate the new pH of the buffer solution using the Henderson-Hasselbalch equation:

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

Substituting the new values,

pH = 4.88 + log (0.496/0.104)

pH = 4.88 + 1.228

pH = 6.11

However, we need to remember that we diluted the original buffer solution by taking a 70.0 mL aliquot and adding 2.00 mL of HI solution.

The new volume of the solution is:

70.0 mL + 2.00 mL = 72.0 mL = 0.0720 L

To account for dilution, we need to recalculate the concentration of sodium benzoate and benzoic acid:

New [A-] = (0.496 mol/L) (70.0 mL/1000 mL) / (0.0720 L) = 0.488 mol/L 
New [HA] = (0.104 mol/L) (70.0 mL/1000 mL) / (0.0720 L) = 0.0102 mol/L

Finally, we can calculate the new pH of the solution:

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

pH = 4.88 + log (0.488/0.0102)

pH = 4.88 + 2.06

pH = 6.94 
However, we need to remember that the question asks for the pH of the 72.0 mL solution, not just the pH of the buffer after adding the HI. Therefore, we need to assume that the added HI and its reaction products are well-mixed in the resulting solution.

In other words, we assume that the 2.00 mL of added HI are uniformly distributed throughout the final 72.0 mL solution.

This means that the total moles of H+ in the solution are:

moles of H+ = (2.00 mL) (2.00 mol/L) = 0.00400 mol

The total moles of buffer components in the solution are:

moles of sodium benzoate = (0.488 mol/L) (72.0 mL/1000 mL) = 0.0351 mol

moles of benzoic acid = (0.0102 mol/L) (72.0 mL/1000 mL) = 0.000734 mol

Therefore, the total moles of H+ and buffer components are:

total moles = 0.00400 mol + 0.0351 mol + 0.000734 mol = 0.0398 mol

The concentration of H+ in the solution is:

[H+] = moles of H+ / total volume of solution

[H+] = 0.0398 mol / 0.0720 L = 0.553 M

And the pH of the solution is:

pH = -log[H+] = -log(0.553) = 0.257 + 0.740

pH = 0.997

Rounded to two significant figures, the pH of the 72.0 mL solution is 1.0.

Therefore, the best choice is D (4.65).

A PH 488 Buffer Was Prepared by Dissolving 0