buffpsoac2a

BUFFER PREPARATION:

1. Prepare 0.5 liters of 0.05 M acetate buffer at pH 5.51.Materials available: 1 M HOAc and solid NaOAc.

Abbreviations for acetic acid: CH3COOH = HOAc and CH3COO Na!  = NaOAc.

"#e p$a of acetic acid is %.&%.

CH3COOH ' === ( CH3COO  ! H! 

). Prepare t#e sa*e buffer usin+ 1 M HCl and solid NaOAc.

3. Prepare t#is sa*e buffer usin+ 1 M HOAc and 1 M NaOH.

%. ,t is desired to c#an+e t#e pH of t#e buffer prepared above to %.-.

  /#ould 1 M NaOH or 1 M HCl be added to t#e eistin+ buffer to do t#is

  Ho2 *an *l of 1 M NaOH or 1 M HCl s#ould be added to t#e eistin+ buffer

ANSWERS:

1. "#e total **ol of bot# for*s of acetate needed = 4500 *l40.05 M = )5 **ol

6ind t#e fraction needed of eac# for*:

 pH = p$ ! lo+ 7A897HA8

5.51 = %.&% ! lo+ 7A897HA8

lo+ 7A897HA8 = 0.&&7A897HA8 = 5.

/o t#e fraction of A = 5.9-. = 0.5 and t#e fraction of HA is 19-. = 0.15

And t#e **ol of A = 40.5)5 = )1.3 **ol.

"#e **ol of HA = 40.15)5 = 3.-) **ol.

;olu*e of 1 M HOAc needed: 4< *l41 M = 3.-)

< = 3.-) *l

ei+#t of NaOAc needed: 4) *+9**ol4)1.3 **ol = 1&50 *+ or 1.&5 +*

"o prepare t#e buffer> one 2ould ta?e a 500 *l volu*etric flas?> fill about #alf full 2it#distilled 2ater> add t#e NaOAc and dissolve> t#en add t#e 3.-) *l HOAc> fill t#e flas? to

t#e *ar? and *i. "#is is t#en 500 *l of 0.05 M acetate buffer at pH 5.51.

). "#is preparation re@uires t#e sa*e ratio and **ol of HA and A .

Ho2ever> all )5 **ole of acetate *ust co*e fro* sodiu* acetate.

ei+#t of NaOAc needed: 4) *+9**ol4)5 **ol = )050 *+ or ).05 +*And 3.-) **ol of t#is *ust be converted to HOAc b addin+ 3.-) *l of 1 M HCl.

3. "#is preparation a+ain re@uires t#e sa*e ratio and **ol of HA and A .

Ho2ever> all )5 **ole of acetate *ust co*e fro* l M acetic acid.;olu*e of l M HOAc needed: 4< *l41 M = )5

< = )5 *l

And )1.3 **ol of t#is *ust be converted to NaOAc b addin+ )1.3 *l of 1 M NaOH.

%. Calculate t#e nu*ber of **ol of A and HA needed for pH %.-. pH = p$ ! lo+ 7A897HA8

%.- = %.&% ! lo+ 7A897HA8

lo+ 7A897HA8 = 0.0-

7A897HA8 = 0.& 

/o t#e fraction of A = 0.&91.& = 0.%-5 and t#e fraction of HA is 191.& = 0.535

And t#e **ol of A = 40.%-5)5 = 11.-)5 **ol.

"#e **ol of HA = 40.535)5 = 13.3&5 **ol.

Ori+inal buffer> pH 5.51 A = )1.3 **ol HA = 3.-) **ol Ne2 buffer> pH %.- A = 11.-)5 **ol HA = 13.3&5 **ol

ifference loss of .&55 **ol +ain of .&55 **ol

HCl is re@uired to convert A to HA.Convert .&55 **ol of A to .&55 **ol HA b addin+ .&55 **ol 1 M HCl.