Develop a recipe for 1.00L of pH 4.50 buffer using only 17.4M HC2H3O2 and 3.00M NaOH (and deionized water). Be sure that it can accommodate an addition of at least .50 mol of strong acid or base with a change in pH less than/equal to 1.
Take 100 mL of the 17.4M HAc and add NaOH to it.
100 mL x 17.4M = 1740 mmoles
...........HAc + OH^- ==> Ac^- + H2O
initial....1740...0........0.......0
added..............x...........
change.......-x...-x.......+x
equil.....1740-x...0........+x
pH = pKa + log (Ac^-)/(HAc)
4.50 = 4.757 + log(x)/(1740-x)
Solve for x and I get about 620 but you need to do it more accurately.
If x = 620 mmoles, then 1740-x = about 1120 mmoles. Convert 620 mmoles NaOH at 3.00 M to mL from M = mmoles/mL.
At this point you should check the HH equation to see that if you substitute the base and the acid into the HH equation that you do in fact obtain pH = 4.50.
Then you want to check if adding 500 mmoles (0.5 mol) acid/base will change the pH more than +/-1. If we take, for example, adding 500 mmoles H^+, it will look like this.
.........Ac^- + H^+ ==> HAc
initial..620....0........1120
add............500............
change...-500..-500......+500
equil...120.....0........1620
Substitute those into the HH equation and I obtained 3.53 but that isn't exact because I've rounded above.
Do the same thing for adding 500 mmoles base to HAc and see that it stays below 5.5.
Post your work if you get stuck.