A volume of 25.0 mL of 0.100 M C₆H₅CO₂H(aq)is titrated with 0.100 M NaOH(aq).What is the pH after the addition of 12.5 mL of NaOH? (K_a of benzoic acid = 6.3 $\times$ 10^-5) A) 2.60 B) 4.20 C) 5.40 D) 7.00 E) 8.60

Question

Quizplus · Accepted Answer

First, we need to write the balanced chemical equation for the reaction between CS1U1B16S1U1B0HS1U1B15S1U1B0COS1U1B12S1U1B0H and NaOH:

CS1U1B16S1U1B0HS1U1B15S1U1B0COS1U1B12S1U1B0H + NaOH ⟶ CS1U1B16S1U1B0HS1U1B14S1U1B0COO1U1B-1S1U1B3S1U1B0Na1U1B+ + H2O

We can see that 1 mole of CS1U1B16S1U1B0HS1U1B15S1U1B0COS1U1B12S1U1B0H reacts with 1 mole of NaOH to produce 1 mole of CS1U1B16S1U1B0HS1U1B14S1U1B0COO1U1B-1S1U1B3S1U1B0Na1U1B+ and 1 mole of water.

Before any NaOH is added, the solution contains only CS1U1B16S1U1B0HS1U1B15S1U1B0COS1U1B12S1U1B0H and water, so the concentration of benzoic acid is 0.100 M.

When 12.5 mL of 0.100 M NaOH is added, the number of moles of NaOH added is:

(0.100 M) x (0.0125 L) = 0.00125 mol

This is also the number of moles of benzoic acid that react, since the reaction is 1:1.

The initial number of moles of benzoic acid is:

(0.100 M) x (0.0250 L) = 0.00250 mol

The number of moles of benzoic acid remaining after the reaction is:

0.00250 mol - 0.00125 mol = 0.00125 mol

The volume of the solution is now 25.0 mL + 12.5 mL = 37.5 mL.

The concentration of benzoic acid after the reaction is:

0.00125 mol / 0.0375 L = 0.0333 M

We can now use the Ka value of benzoic acid to calculate the pH:

Ka = (H+)(C6H5COO-)/(C6H5COOH)

6.3 x 10^-5 = (H+)(0.0333)/(0.100)

(H+) = 1.90 x 10^-5 M

pH = -log(H+) = 4.72

However, we still need to account for the fact that we added NaOH, which will react with water to produce OH- ions. We can use the concentration of NaOH added to calculate the concentration of OH- ions produced:

0.100 M NaOH x 0.0125 L = 0.00125 mol OH-

The volume of the solution is now 37.5 mL, so the concentration of OH- ions is:

0.00125 mol / 0.0375 L = 0.0333 M

pOH = -log(0.0333) = 1.48

pH + pOH = 14, so:

pH = 14 - pOH = 14 - 1.48 = 12.52

However, this calculation assumes that all of the added NaOH has reacted with water to produce OH- ions. In reality, some of the OH- ions will react with the remaining benzoic acid to produce the benzoate ion (C6H5COO-). We can use the Henderson-Hasselbalch equation to account for this:

pH = pKa + log([C6H5COO-]/[C6H5COOH])

pKa = -log(Ka) = -log(6.3 x 10^-5) = 4.20

[C6H5COO-]/[C6H5COOH] = 0.0333/0.100 = 0.333

pH = 4.20 + log(0.333) = 4.72

Therefore, the pH after the addition of 12.5 mL of NaOH is 4.72, which corresponds to answer choice B.

A Volume of 25 ×\times× 10-5) A) 2

A Volume of 25 $\times$ 10^-5)
A) 2