# Data and calculations Standarization of silver nitrate Standard NaCl titration trials Trial

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

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

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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