15.25 grams of acetic acid solution (density = 1.00g/mL) required 16.05mL of 0.875M sodium hydro

Question

Snehal J · Accepted Answer

1. First, we need to find the moles of sodium hydroxide (NaOH) used in the titration. We can do

15.25 grams of acetic acid solution (density = 1.00g/mL) required 16.05mL of 0.875M sodium hydroxide solution to reach the endpoint of a titration. Using this data, the acetic acid would have a concentration of M. (round answer to 3 significant figures)