Question

1. Calculate the pH of the acetate buffer, mmol. HA, mmol. A-, and mmol. H3O+ after 4mL, 8mL, 12mL, 16mL, 20mL, 24mL, and 28mL of 0.100M HCl is added to the acetate buffer.

2. Calculate the pH of the acetate buffer, mmol. HA, mmol. A-, and mmol. H3O+ after 4mL, 8mL, 12mL, 16mL, 20mL, 24mL, and 28mL of 0.100M NaOH is added to the acetate buffer.

3. Calculate the pH of the ammonia buffer, mmol. HA, mmol. A-, and mmol. H3O+ after 4mL, 8mL, 12mL, 16mL, 20mL, 24mL, and 28mL of 0.100M HCl is added to the ammonia buffer.

4. Calculate the pH of the ammonia buffer, mmol. HA, mmol. A-, and mmol. H3O+ after 4mL, 8mL, 12mL, 16mL, 20mL, 24mL, and 28mL of 0.100M NaOH is added to the ammonia buffer.

Given:

0.160M NaC2H3O2 / 0.200M HC2H3O2

0.160M NH4Cl / 0.200M NH3

0.100M NaOH

0.100M HCl

Volume of Buffer: started by adding 10.00mL buffer with 30.0 mL distilled water

Please help! I know it is a lot of calculations, but even if you could just show my the steps for a few of the questions it would be a great help!

Answer 1

for acidic Buffer, pt= pka + log [salt] land] sme c we have [Chorona 3 = o.16OM P [ CHz Cook] = 0.200M 0.000 form : P = 4.74 + log fets Cowa] [CHS2004] - 4.74 + bg 0.160 8.200 = 4.64 Total volume of Buffer 40mt Butter contains (10ml + 30 ml) = Gome. 40x0.160 o mol Cts Loo ion 1000 - 0.0064 moh Ct₂coo-ion, 40m Butter contains 404 0.200 mot city cooh 1000 = 0.008 mol cits cooh. 1000 0.0004 when 4 ml o.L M HCl added then 9x0.1 mol itt combines with o.o004 mol che coo-ion to form 0.0004 mol Cts CoolH. : After addition of a ne ol (M) Hal- No. of moles of Checoo- = 0.0064 -0.0004 mol = 0.0060 mol No. of mor. of Chi Colt = 0.0080+ 0.0004 mil = 0.0084 mol. moles of alts coo- r since volume pH = 4.74 + bog? I is same moles of its coolt! so direct use of - 4.74 + 0.00 60 moles gives 70'0084 the same - 4.59 result

me o.1 (M) He contains 8 mt xo.lM 1000m - 0.0008 mol Ations. 0.0008 mot Ht combines with 0.0088 mer che coon to form 0.0008 mot ut 400H. CH CHO & H t - its cooh 0.0008mel. 0.0008 mal 8.0oogmor Alter addition of sml 0.IM HU- No. of moles of at coo- = 0.0064-0.0008 mot = 0.0056mos. On of Cits cooH = 0.0080 + 0.0008 mot - 0.0088 mol. - pt - 4.74 + bg oo 0.0056 0'0088 = 4.54. 12 ml ont (M) Al contains 12m1 XO 1M = 0.0012 mot 1000 ml + 0.0012 mot Ht combines with 0.0012 mot etsuro to form 0.00 12 mol cits cool. atter addition of $2.0m2 o.tm Al- No. of molesot Cts coo- = 0.0064 -0.0012 mor -0.0052 mol. No. of moles of the cooH = 0.0080 +0.0012 mof - 0.0092 mol. 0.0052 PH = 4.74 + log 8.0092 - 4.49

NO @ pH of the acetate buffer = 4.64. when 4 mt 0.100 M Naot added. Then ho.cf moles 4 x 0.1 1000 1 = = 0.0ooumor. 0 OH " 0.000 y mol ott non nutralise sooou motetscoot orm o.oooumot. It coorion. & Astor addition of 4mL ont (M) Natha No. of mores of Cty coo- = 0.00 64 +0.0004 mots = 0.0068 mots No. of moles of the coolH = 0.0088 -0.0004 mots - 0.0076 mols. [0.0068 pH = 4.74 + log 0.0076 - 4.69 Similar calculation for 8 mL, 12mt, 16 mt 20m2, 24me and 28 me addition. when 28ml 0.1 (M) Nadh added, then no. of moles of oth added = 28x0.1 1006 - 0.0028 mot. 0.00 28 moles OH ion neutralise d. 0028 moles CHS COOH to form 0.0028 mots of city coo- ion. . Nooit son moles of the coor = 0.0064 + 0.0028 - 00.0092 moles,

not moles of its coolt = 0.0080 - 0.0028 -0.0032 mots. soles pH = 4.74 + log 0.00012 = 4.99 0.00 52 *Cook mots pt of a amonia buffer is givenby pH = 14- (pkb + boy Esalt] ) Clase] . we have set (Na 43 = 0.1601 Imots [ NH 320.200M. :: pl = 14- (4.75 + by 2020. ). = 9.35. Now, when & me ont (M) He added the no.uz moles of -lt + added 4 ne y oid 5=0.000 mar. 1000 me 0.0004 mol Ht combines with 0.000 4 mot Nits to form 0.000 y mot Nhat. : No. of moles of Nitet = 0.0064 +0.00 ou met = 0.006 8 mot. No. of moles of NH = 10.0080 -0.0004 met = 0.0076 mot. .PH _ 14- (4.75 + by o. 0068 2 9.29 Similar calculation for other addition. 0.0076)

o pH of ammonia buffer= 9.35. Now. 4 ml oil (M) NaOH added to the buffer solution. 4 ml xoll :. No. of moles of ot-added - 1000 ml 2 0.0004 mo. Now, nomi Bufker lon tains - 40 ml xo.160 1000 me = 0.0064 mots Nitat 90 met 0.200 1000 me - 0.0080 mots. A what 0.000 4 mots of reacts with what to form 0.0004 mots of NHgGH for Nits). After addition of 0.0004 mets ot- No. of moles of Nihat = 0.0064 -0.0004 20.0obo mols No. of moles of Wit = 0.0080 + 0.0004 20.0084 mois p4_10- (4.85€ logogu) 0.0060 9.39 Similar calculations for sthus additions,