# Answered Essay: Fill out the whole table, you do not have to show your work on all of them, just show

Fill out the whole table, you do not have to show your work on all of them, just show your work on 4 different kinds, example:

Problem #1: an acid that I measured the pH, we turn that into hydrogen ion concentration and fill out the table.
Problem #2: an acid where I did not measure the pH, in which case you start by looking up kA, then fill out the table
Problem #3: a base where I measured the pH turn that into hydroxide concentration and fill out the table,
Problem #4: a base where I did not measure the pH, then you have to find kB and fill out the table.

You only have to show work for each one of these and the rest can be done without any work.

In that table if you were to look at “Exp pH” you could see that some of them are crossed over and another pH is written, those ones are called “measured” for the measured columns you have to fill out “Theo Ka or Kb” and the “%error” as well. For the non measured rows, you do not have to fill out “Theo Ka or Kb” and the “%error.”

NOTES:
1) For the Conjugate Acid and Conjugate Base, do not put H+, OH-. If it’s a weak acid, put that weak acid. For Conjugate base, then put the weak base. If the compound was a weak base, then put that thing in there, then put the conjugate acid in the other column. And do not include any spectator ions.

2) Do not bother to use the ice table to determine X, values for Ka or Kb can be looked up online.

3) In the case of nitrate ion, we cannot Kb, we can only look up Ka for its conjugate acid.

4) 7 of them have their pH measured. And the other 7 do not and that should be looked up.

Examples

(i) pH = 2.54 = -log[H+]

[H+] = antilog (-pH )= 0.00288 M

(ii) Ka = [H+][A-] / [HA]

Ka = x2 / [HA]

x = [H+]

pH = 9.14 = -log[H+]

[H+] = 7.244 X 10-10 M

Ka = (7.244 X 10-10)2 / 0.25 = 2.1 X 10-20

(iii) pOH = 14 – pH

pOH = 14- 10.96 = 3.04

(iv) [OH-] = 10^-14 / [H+] =

pH = -log [H+] = 5.88

[H+] = antilog(-[H+]) = 1.32 X 10-6 M

[OH-] = 7.58 X 10-9 M

(v) Kb = Kw / Ka = 10-14 / 7.244 X 10-10 = 1.38 X 10-5

(vi) pKb = -logKb = 4.86

(vii) % dissociation = [H+] X 100 / [HA] =

[H+ ]= 0.00288

% dissociation = 0.00288 X 100 / 0.25 = 1.152%

 0.252 Exp pH minus pH Conjugate acid [H+] Ka pKa Conj base pOH minus pOH [OH-] Kb pKb % dissociation Theo pKa Theo pKb TheoKa or Kb %error of K CH3NH2 11.95 -11.95 CH3NH2+ 1.12202E-12 2.05 -2.05 0.008912509 0.00031521 3.501401 3.536710072 HC2H3O2 2.54 -2.54 0.002884032 3.30065E-05 4.481401 C2H3O2- 11.46 -11.46 3.46737E-12 1.144456946 4.75 1.77828E-05 5.654725 NaC2H3O2 9.14 -9.14 H2C2H3O2 7.24436E-10 4.86 -4.86 1.38038E-05 7.56135E-10 9.121401 0.005477715 NH3 10.96 -10.96 NH4+ 1.09648E-11 3.04 -3.04 0.000912011 3.30065E-06 5.481401 0.361909063 NH4+ 5.88 -5.88 1.31826E-06 6.89604E-12 11.1614 NH3 8.12 -8.12 7.58578E-09 0.000523118 9.25 5.62341E-10 -20.6638 NaHSO3 1.36 -1.36 0.043651583 0.007561352 2.121401 NaSO3- 12.64 -12.64 2.29087E-13 17.32205683 H2SO3 1.17 -1.17 0.067608298 0.01813842 1.741401 HSO3- 12.83 -12.83 1.47911E-13 26.8286895 NA2CO3 11.54 -11.54 NaHCO3 2.88403E-12 2.46 -2.46 0.003467369 4.77089E-05 4.321401 1.375939883 NaHCO3 9.5 -9.5 H2CO3 3.16228E-10 4.5 -4.5 3.16228E-05 3.96825E-09 8.401401 0.012548721 H3PO4 1.6 -1.6 0.025118864 0.002503799 2.601401 H2PO4- 12.4 -12.4 3.98107E-13 9.9678033 KH2PO4 4.44 -4.44 3.63078E-05 5.23118E-09 8.281401 KHPO4- 9.56 -9.56 2.75423E-10 0.014407859 K3PO4 12.86 -12.86 K2HPO4 1.38038E-13 1.14 -1.14 0.072443596 0.020825693 1.681401 28.74745873 K2HPO4 10.25 -10.25 KH2PO4 5.62341E-11 3.75 -3.75 0.000177828 1.25487E-07 6.901401 0.070566643 C18H21NO3 10.91 -10.91 C18H22NO3+ 1.23027E-11 3.09 -3.09 0.000812831 2.6218E-06 5.581401 0.322551792 H3BO3 4.98 -4.98 1.04713E-05 4.3511E-10 9.361401 H2BO3- 9.02 -9.02
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