You want 0.075 mols to start HAc. You're right there (and the 4.50 g is correct).
5.00 = 4.75 + log (b/a)
b/a = 1.82 or
b = 1.82*a or
a = 0.549*b. Substitute this into the below for a.
b+ a = 0.15 x 0.5 = 0.075 mols.
b + 0.549b = 0.075
1.549b = 0.075
b = 0.0484 mols.
3.0M = 0.0484 mols/L
and L = 0.0161 L or 16.1 mL.
Check my arithmetic.
You are asked to prepare 500. mL of a 0.150 M acetate buffer at pH 5.00 using only pure acetic acid, 3.00 M NaOH, and water. Calculate the quantities needed for each of the following steps in the buffer preparation.
1. Add acetic acid to ~400 mL of water in a 500 mL beaker. How many grams of acetic acid do you need?
2. Add 3.00 M NaOH solution while stirring and monitoring the pH with a pH electrode until the pH is 5.00. What volume of NaOH solution do you calculate that you will need?
By taking the desired moles of the solution (0.15M x 0.5L = 0.075 moles), I was able to calculate the needed Acetic Acid (Question #1 = 4.5g)
(Desired Moles) 0.075 x (Formula Weight) 60.05 = 4.5g Acetic Acid.
Where I am stuck is Question 2. After using HH:
(AC/HAC)=10^(5-4.76), (AC/HAC)=1.74
After this, I multiplied my HAC (0.075) by my 1.74 to achieve my AC (0.13).
I know I need to incorporate this with my Molarity of base (3.00M), but am not sure how. Am I even on the right track?
Thank you very much in advance.
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