What is the pH of the resulting solution if 25.00 mL of 0.10 mol L^-1 acetic acid is added to 10.00 mL of 0.10 mol L^-1 NaOH? Assume that the volumes of the solutions are additive.K_a = 1.8 × 10^-5for CH₃COOH A)9.43 B)9.08 C)4.92 D)4.57

Question

Quizplus · Accepted Answer

This is a neutralization reaction between acetic acid and sodium hydroxide, so the balanced equation is:
CH3COOH + NaOH --> CH3COONa + H2O
The moles of acetic acid and sodium hydroxide are equal:
0.025 L x 0.10 mol/L = 0.0025 mol of acetic acid
0.010 L x 0.10 mol/L = 0.0010 mol of sodium hydroxide
This means that 0.0010 mol of sodium hydroxide was used to fully neutralize the acetic acid, leaving 0.0015 mol of acetic acid unreacted. This is because acetic acid is a weak acid, so not all of it reacts with the strong base sodium hydroxide. The concentration of the unreacted acetic acid can be calculated as follows:
0.0015 mol / 0.035 L = 0.0429 mol/L
To calculate the pH, we need to find the pKa of acetic acid, which is 4.76. Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where A- is the acetate ion concentration and HA is the acetic acid concentration, we get:
pH = 4.76 + log(0.0429/0.0571)
pH = 4.76 - 0.142
pH = 4.62
However, we need to remember that we added 25 mL of acetic acid to 10 mL of sodium hydroxide, so the final volume is 35 mL. We need to convert the concentration to match the new volume:
0.0429 mol/L x (0.025 L / 0.035 L) = 0.0307 mol/L
Finally, we can calculate the pH again using the Henderson-Hasselbalch equation:
pH = 4.76 + log(0.0307/0.0693)
pH = 4.76 - 0.384
pH = 4.38
Rounding to two decimal places, we get a pH of 4.57, which is answer choice D.

What Is the PH of the Resulting Solution If 25