Data and calculations
Standarization of silver nitrate
Standard NaCl titration trials
Trial
NaCl (weight)
End Point
1
0.2377 g
13.83 mL
2
0.2430 g
14.00 mL
3
0.2563 g
14.70 mL
Titrant (AgNO3) added volume (Vt):
Trial 1
Vt = 13.83 mL
Trial 2
Vt = 14.00 mL
Trial 3
Vt = 14.70 mL
Reaction
NaCl + AgNO3 AgCl (s) + NaNO3
1 mol NaCl = 1 mol AgNO3
NaCl moles calculations (MW NaCl = 58.44 g/mol)
Trial 1
Mol NaCl = (0.2377 g) x (1mol/58.44g) = 4.0674 mmol NaCl = mmol AgNO3
Trial 2
Mol NaCl = (0.2430 g) x (1mol/58.44g) = 4.1581 mmol NaCl = mmol AgNO3
Trial 3
Mol NaCl = (0.2563 g) x (1mol/58.44g) = 4.3857 mmol NaCl = mmol AgNO3
Real [AgNO3]
[AgNO3] = Mol / Vt
Trial 1
[AgNO3] = (4.0674 mmol AgNO3)/(13.83 mL) = 0.2940 M
Trial 2
[AgNO3] = (4.1581 mmol AgNO3)/(14.00 mL) = 0.2970 M
Trial 3
[AgNO3] = (4.3857 mmol AgNO3)/(14.70 mL) = 0.2983 M
Mean [AgNO3]
Mean [AgNO3] = ([AgNO3]1 + [AgNO3]2 + [AgNO3]3 + [AgNO3]4) / 3
[AgNO3] = (0.2940 M + 0.2970 M + 0.29283 M) / 3 = 0.2964 M
Determination of an unknown
Sample
Unknown weight
End Point
1
0.4279 g
22.93 mL
2
0.3447 g
18.30 mL
3
0.4002 g
21.10 mL
Titrant (AgNO3) added volume (Vt):
Trial 1
Vt = 22.93 mL
Trial 2
Vt = 18.30 mL
Trial 3
Vt = 21.10 mL
Reaction
Cl- + AgNO3 AgCl (s) + NO3-
1 mol NaCl = 1 mol AgNO3
Moles AgNO3 added ([AgNO3] = 0.4718 M)
Moles AgNO3 = [AgNO3] x Vt
Sample 1
Moles AgNO3 = (0.2964 M) x (22.93 mL) x (1 L/1000 mL) = 6.7964 mmol AgNO3 = mmol Cl-
Sample 2
Moles AgNO3 = (0.2964 M) x (18.30 mL) x (1 L/1000 mL) = 5.4241 mmol AgNO3 = mmol Cl-
Sample 3
Moles AgNO3 = (0.2964 M) x (21.10 mL) x (1 L/1000 mL) = 6.2540 mmol AgNO3 = mmol Cl-
Mass Cl- (AM Cl = 35.5 g/mol)
Sample 1
Mass Cl- = (6.7964 mmol Cl-) x (1 mol/1000 mmol) x (35.5 g/mol) = 0.2412 g Cl-
Sample 2
Mass Cl- = (5.4241 mmol Cl-) x (1 mol/1000 mmol) x (35.5 g/mol) = 0.1925 g Cl-
Sample 3
Mass Cl- = (5.9576 mmol Cl-) x (1 mol/1000 mmol) x (35.5 g/mol) = 0.2220 g Cl-
% Cl-
% Cl- = [(g Cl-) / (g sample)] x 100%
Sample 1
% Cl- = [(0.2412g) / (0.4279 g)] x 100% = 56.37%
Sample 2
% Cl- = [(0.1925g) / (0.3447 g)] x 100% = 55.84%
Sample 3
% Cl- = [(0.2115g) / (0.4002 g)] x 100% = 55.48%
Mean % Cl-
Mean % Cl- = [(g Cl-)1 + (g Cl-)2 + (g Cl-)3 ] / 3
Mean % Cl- = [ 56.37% + 55.84% + 55.48%] / 3 = 55.90%
Part 3
Sample
Mass
Crucible filter
Crusible filter w/AgNO3
AgNO3
1
0.4148 g
30.2707 g
31.2109 g
0.9402 g
2
0.4095 g
31.0544 g
31.9894 g
0.9350 g
3
0.4053 g
30.7135 g
31.6364 g
0.9220 g
4
04103 g
30.7341 g
31.6651 g
0.9310 g
Mass Cl- (AM Cl = 35.5 g/mol, MM AgNO3 = 143.32 g/mol))
Mass Cl- = (Mass AgNO3) x (1 mol AgNO3/143.32 g AgNO3) x (1 mol Cl-/1 mol AgNO3) x (35.45 g/molCl)
Sample 1
Mass Cl- = (0.9402 g) x (1 mol AgNO3/143.32 g AgNO3) x (1 mol Cl-/1 mol AgNO3) x (35.45 g/molCl) = 0.2326 g Cl-
Sample 2
Mass Cl- = (0.9350 g) x (1 mol AgNO3/143.32 g AgNO3) x (1 mol Cl-/1 mol AgNO3) x (35.45 g/molCl) = 0.2313 g Cl-
Sample 3
Mass Cl- = (0.9220 g) x (1 mol AgNO3/143.32 g AgNO3) x (1 mol Cl-/1 mol AgNO3) x (35.45 g/molCl) = 0.2283 g Cl-
Sample 4
Mass Cl- = (0.9310 g) x (1 mol AgNO3/143.32 g AgNO3) x (1 mol Cl-/1 mol AgNO3) x (35.45 g/molCl) = 0.2303 g Cl-
% Cl-
% Cl- = [(g Cl-) / (g sample)] x 100%
Sample 1
% Cl- = [(0.2326g) / (0.4148 g)] x 100% = 56.08%
Sample 2
% Cl- = [(0.2313g) / (0.4095 g)] x 100% = 56.48%
Sample 3
% Cl- = [(0.2283g) / (0.3447 g)] x 100% = 56.58%
Sample 4
% Cl- = [(0.2303g) / (0.4103 g)] x 100% = 56.13%
Mean % Cl-
Mean % Cl- = [(g Cl-)1 + (g Cl-)2 + (g Cl-)3 + (g Cl-)4] / 4
Mean % Cl- = [56.08% + 56.48% + 56.58% + 56.13%] / 4 = 56.32%
Reference:
Harris, D.C., Lucy, C.A., Quantitative Chemical Analysis, 10th Ed; W.H. Freeman and Company: NY, 2020
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