If 25 mL of 0.20 M NaOH is added to 50 mL of 0.10 M HC₂H₃O₂ (Ka = 1.8 × 10^-5), what is the pH? A) 5.33 B) 10.21 C) 8.78 D) 13.56 E) 1.34

Question

Quizplus · Accepted Answer

First, we need to write out the balanced equation for the reaction of HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0 (acetic acid) with NaOH:
HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0 + NaOH → NaCS1U1B12S1U1B0HS1U1B13S1U1B012S1U1B0 + H2O

Next, we need to determine how much of each species is present before the reaction occurs. We have 50 mL of 0.10 M HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0, which is equivalent to 0.005 mol HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0. We also have 25 mL of 0.20 M NaOH, which is equivalent to 0.005 mol NaOH.

Since NaOH is a strong base, we can assume that it will completely react with acetic acid to produce its conjugate base, sodium acetate. Therefore, after the reaction occurs, we will have 0.005 mol NaCS1U1B12S1U1B0HS1U1B13S1U1B012S1U1B0 and 0 mol HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0 remaining.

Now, we need to calculate the concentration of each species in the final solution, which we will use to determine the pH. We know the total volume of the solution is 75 mL (50 mL + 25 mL), so we can calculate the total moles of solute:
0.005 mol NaOH + 0.005 mol HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0 → 0.005 mol NaCS1U1B12S1U1B0HS1U1B13S1U1B012S1U1B0

The final volume is also 75 mL, so the concentration of the sodium acetate is:

C(NaCS1U1B12S1U1B0HS1U1B13S1U1B012S1U1B0) = 0.005 mol / 0.075 L = 0.067 M

Now, we can use the equilibrium expression for the dissociation of acetic acid to calculate the concentration of H+ ions in solution:

Ka = [H+][C2H3O2-] / [HC2H3O2]

We know the initial concentration of acetic acid and the final concentration of the conjugate base, so we can solve for [H+]:

Ka = 1.8 × 10S1U1P1-5S1S1P0 = x^2 / (0.005 - x)

Simplifying this expression and using the quadratic formula, we find that:

[H+] = 3.3 × 10S1U1P1-9S1S1P0 M

Finally, we can calculate the pH:

pH = -log[H+] = 8.78

If 25 ML of 0