A scientist makes 300.0 mL of an acetate buffer using [CH3COOH]o=[NaCH3COO]o= 0.075 M. How many mL of 5.0 M NaOH can be added before its buffering capacity is exhausted?
You have (HAc) = A = 0.075M
and (Ac^-) = B = 0.075M
you want to add "x" L of 5.0M to 300 mL buffer to produce a pH of pKa + 1 or 4.76 + 1.00 = 5.76.
........A + OH^- ==> B + H2O
I....0.075..0.....0.075.....
add.........x................
C......-x..-x........+x
E..0.075-x..0.....0.075+x
5.76 = 4.76 + log(0.075+x)/(0.075-x)
Solve for x = additional OH^- needed. I found about 0.062M but you need to confirm that and watch the significant figures while you're at it.
Therefore, for 1000 mL of this solution you will need to add 0.062 mol base. Since M = mols/L, substitute and solve for L. I get about 0.012 L or 12 mL (for 1L solution) which is 12 x (300/1000) = about 3.7 mL of 5M base. Again, that 3.7 is not exact. You need to go through the problem. I like to check these things to see if we in fact changed the pH by 1.00 unit when we added the base.
We had 300 mL x 0.075 = 22.5 mmoles A and 300 mL x 0.075 = 22.5 mmoles B.
We add 5M x 3.7 mL = about 18.5 mmols.
The acid is now 22.5-18.5 = about 4 and
base is now 22.5+18.5 = about 41 so
pH = 4.76 + log(41/4)
pH = 5.77 which probably will be closer to 5.76 if you go through more exactly than I did. Check my work.