Author Question: What is the pH of the resulting solution if 40.00 mL of 0.10 M acetic acid is added to 10.00 mL of ... (Read 53 times)

darbym82 · **on:** Feb 15, 2020

Question 1

Formic acid (HCO2H, Ka = 1.8 × 10-4) is the principal component in the venom of stinging ants. What is the molarity of a formic acid solution if 25.00 mL of the formic acid solution requires 44.80 mL of 0.0567 M NaOH to reach the equivalence point?
◦ 0.0134 M
◦ 0.0316 M
◦ 0.102 M
◦ 0.0567 M

Question 2

What is the pH of the resulting solution if 40.00 mL of 0.10 M acetic acid is added to 10.00 mL of 0.10 M NaOH? Assume that the volumes of the solutions are additive. Ka = 1.8 × 10-5 for CH3CO2H

◦ 8.78
◦ 4.27
◦ 5.22
◦ 9.73

kingdude89 · **Reply #1 on:** Feb 15, 2020

A FREE account is required to view all solutions!

Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.

Answer Preview

Only 39% of students answer this correctly