Question

Given: K for acetic acid is 1.8 X 10–5You are titrating 0.108 M NaOH into 142.0 ml of acetic acid of unknown concentration. You have an indicator that will change color when equivalence is reached. At equivalence, you have added 72.0 ml of the base. Calculate molarity of the acid. What is the pH of the solution at the equivalence point? Now that you know the molarity of the acid, find pH when you mix 50.0ml of the acid with 75.0 ml of the same NaOH solution. Now you are working with different acid and base, both weak. K for the acid is 2.25 X 10-5. You mix 63 ml of 0.275 M acid with 55.0 ml of a weak base of concentration 0.188 M. Find pH

Answer 1

Answer:

Explanation:0.493 M NaOH means 0.493 mol NaOH/L

mols

mols = ------ x L

L

mols = M x V

In a titration procedure, 40.57 mL of 0.493 M NaOH solution was used. How many mols NaOH did this volume of NaOH solution contain?

mols = M x V

0.493 mols NaOH

mols = ----------------------- x 0.04057 L

L

mols = 0.0200 mols NaOH