Question

Say for example, that you had prepared a Buffer C, in which you mixed 8.203 g of sodium acetate,NaC2H3O2, with 100.0 mL of 1.0 M acetic acid.

1A. What would be the initial pH of Buffer C

1B. If you add 5.0 mL of 0.5 M NaOH solution to 20.0 mL each of Buffer B( initial pH of Buffer B is 3.83), and Buffer C, Which buffer's pH would change less? Explain

Answer 1

Thus change in pH of buffer B is more. Without doing calculations this could also be answered with the knowledge that a buffer has maximum buffer capacity when pH of the buffer is equal to pKa of the acidic part and here buffer C had initially pH equal to pKa. So change in pH of C would me less and that of B would be more.

(A) pH of a buffer can be found using Henderson - Hasselbalch pH = praf 3 [Acetate] (Acetic Acid] Now, 8.2039 Nach, Co0= 8.2039 = 0.6 mol CHE COONa formula mass 182.03 g mal so, [Cręcoona) = 0.1 mol = 1M pH using equation Ĝi above would be: pH = 4.745 + log ( 1 ) = 4.745 t ein = 14.745 Answer OIL (pka z doy chia io)) (pka = -logka . so = 14.745 Answer acetic and (B) In 20ml, total moes of acetic acid plus acetate = 20 times total moles in loom Joo“ a loimo to.s mot)= 0.2 = 0.04 mot In buffers moes naoh added = 0.5Mx 5 xol = 2.5x15 mal -0.0025 mol moles of autate now = 0.07+0.0025 0.025 mol Initial moles of acetate= 0.02 tacetic acid converted to acetate I by reaction with added NaOH = 0.02-0.0025 = 0-075 mol: pH = 4.745 + log (9.67354) 4.745 . change in pH= 0109 where Vis total volun! For buffer B B Initially before NaOH addition: pH = pkp + log cacetate) Cácetate] .83-4.745) Jacetic and ] . Tacetic acid] 3.22 4745 = 0.1216 - • Also, acetic acidt acetate moles=o ou = aid acidt o.12 16 acetic acid=0.04 = acetic and= 0.0357 mil (using ID ) acetate = 0.04-0.0357= 0.00434nul Alar 0-0025 Wal NaOH addition, acetic and = 0.0357-0.0025 20.0332 acetate = 0.00684 mel: pH=4.745't log 10.0068ycvi change in pH= $4.059--3-83 = 0.22 0.033 2NS = 4.059 4.745 + log 10.01751 +0109

Comment in case of any doubt.