Can someone help me with the calculations especially with finding pKa1&2 Ka1&2,

ID: 1058539 • Letter: C

Question

Can someone help me with the calculations especially with finding pKa1&2 Ka1&2, the experimental and literature, just getting started would help me. Thank you The Pvkes E01 annizca a2Luin IT, mule o Lusin ok KHaS Undnor AC duatin) 2ami- ratindA))li ACwa ILY ived f-40L Cbfactuie) (a)mol) Pra E ( CrDenn mial,goth) tal ( L la our e) ta 1 ( expenmrhan (Iroaue ) Kaz ( e ennema1) a) taz ( litattur Into! n Hai (calculated) pacev, arr, nitial pH Ctry' aror, Initial pH wakStation Numer- FoopDic Acid O-O Dioric Acid 35.5 FAcrue Had for 4drseccrd par-e-cc reel fdu par tte ,,HRoweet e Had for the second H4career id not cunt 4 art25 04 DATE HAYDENwNER STUDENT LAB NOTEBOOK NOTE ISERT DICER UNDER COPSET BEFORE watNG

Explanation / Answer

Monoprotic acid

Diprotic acid

Volume of acid used (mL)

25.0

20.0

Volume of NaOH used (mL) = (final volume – initial volume)

22.4

35.5

Molarity of NaOH (mol/L)

0.100

Moles of NaOH at equivalence point (moles) = (molarity of NaOH)*(Volume of NaOH in L)

0.00224

0.00355

Moles of acid (mol)

0.00224*1

0.001775*2

Molarity of acid (mol/L) = (moles acid)/(volume of acid in L)

0.0896

0.08875

(Couldn’t figure out this one) (g/L)

16.5771

10.7403

Molecular weight of acid (g/mol) (experimental) = (upper column)/(molarity of acid)

185.0123

121.017

Molecular weight of acid (g/mol) (literature)

164.15

118.09

Percent error MW (%) = (experimental – actual)/actual*100

12.71

2.48

pKa1 (experimental,graph)

3.624*3

pKa1 (literature)

4.0

4.207

Ka1 (experimental)

2.375*10-5

pKa2 (experimental, graph)

pKa2 (literature)

Ka2 (experimental)

Ka2 (literature)

Initial pH (experimental)

3.93

2.16

Initial pH (calculated)

Percent error, initial pH

*1 Write down the neutralization reaction of the monoprotic acid (HA) with NaOH as

HA (aq) + NaOH (aq) ------> Na+A- (aq) + H2O (l)

Moles NaOH added at end point = moles of HA neutralized.

*2 Write down the neutralization reaction of the diprotic acid (H2A) with NaOH as

H2A (aq) + NaOH (aq) -----> Na2A (aq) + H2O (l)

Moles of NaOH added at end point = (Moles of H2A neutralized)/2 (there is a 2:1 molar ratio). I assumed you are talking about complete neutralization; a diprotic acid has two equivalence points corresponding to two dissociations and Ka1 and Ka2.

*3 At the equivalence point, the major species is A- (conjugate base of HA)which establishes equilibrium as

A- (aq) + H2O (l) <====> HA (aq) + OH- (aq)

Since OH- is formed, we must work with pOH. Given pH at the equivalence point, we can calculate pOH as pH + pOH = 14

===> 8.1492 + pOH = 14

===> pOH = 14 – 8.1492 = 5.8508

Therefore, [OH-] = 10-pOH = 10-5.8508 = 1.4099*10-6 M

At the equivalence point, [OH-] = [HA] = 1.4099*10-6 M and moles A- = moles of NaOH added = 0.00224 mol.

Total volume of solution at end-point = (25.0 + 22.4) mL = 47.4 mL.

[A-] = (0.00224 mol)/(47.4/1000) L = 0.0472 mol/L

Kb = [HA][OH-]/[A-] = (1.4099*10-6)2/(0.0472) = 4.211*10-11

Ka = Kw/Kb = 1.0*10-14/(4.211*10-11) = 2.375*10-4)

pKa1 = -logKa1 = -log(2.375*10-4) = 3.624