Question

Describe how would you prepare 500.0 mL of buffer with pH = 9.08 and [HA] + [A−] = 0.10 M using the following reagents available to you in the stock room and any appropriate glassware.

Available:

NH4Cl (F.W. = 53.5)   Ka = 5.5×10−10 pKa = 9.26

6.00M HC2H3O2       Ka = 1.8×10−5    pKa = 4.74

NaH2PO4•H2O (F.W. = 137.99)       Ka = 7.4×10−8    pKa = 7.13

3.00M NaOH            

Deionized H2O

1) Pick the correct weak acid.

2) Solve the Henderson-Hasselbalch equation for the ratio of [base]/[acid].

3) Find the [base] and [acid]. (easiest using two equations and two unknowns)

4) Find the initial concentrations of reagents. It may help to set up an ICE box using the [base] and [acid] found in 3 as the equilibrium values, and work in reverse.

5) Convert the initial concentrations into grams for solids and mL for solutions.

Answer 1

9.08 = Tacid) For preparation of a buffer, we select an acid with pka close to required pH. (a) so Acid selected = NHut ,ie. Nhull pka=9.26. pH = pka + log [C.B.] Cacid] 1.26 tlog CC. Bol so log [C.B.] = 9.08-926= -018 (C-B] o lacid] = 10-0.18 [aud] = 0.66 Now base = 0.66 and base ta cid=0.10 aud so base = 0.66 acid Oo66 acid & auid = 0olo 1066 acid = 0.10 a cid=6602X10-2M base = 0.66 X 6.02 X 10 M = 3.97 x 10-2M

Initial moles of NHyų added = Molarity x volun - Oolo molly x Oost = 0.05 mol Moles of NH₃ required = 3.97X10-2 mol txo.5% = 1.985. 10-2 mol Nhout Noon - ace + NH3 + HO I mol Nao give Imol NH so 1.985 * 10-2 mol is obtained by using 10985 10-2 mol Naon. Basic procedure ) Dissolve 0.05 mol of NHy l in 5oo.omL water and add 1.9858 10-2 mol Naon. low Mass of which required = No of moles x FW = 0.05 mot X 53-5 gmot = 2.6759. volume of Naon required = No. of moles = 1.9858 10 mol = 6.62mL. olarity 3.00 mbl L- so Dissolve 206759 Nhull in minimum amount of water (about 40 - somL). Add dropwise 6.62mL of 3.00 M NaOH and make up the volume 500.0 mL by adding Di water. ce Now