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Determine the pH during the titration of 75.0 mL of 0.100 M HCl by 0.100 M NaOH … ( a ) before the addition of any NaOH. We have 0.100 M HCl in the flask. pH = -log(0.100) = 1.000. Determine the pH during the titration of 75.0 mL of 0.100 M HCl by 0.100 M NaOH …

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Determine the pH during the titration of 75.0 mL of 0.100 MHCl by 0.100 MNaOH…

(a) before the addition of any NaOH.

We have 0.100 M HCl in the flask.

pH = -log(0.100) = 1.000

Determine the pH during the titration of 75.0 mL of 0.100 MHCl by 0.100 MNaOH…

(b) after adding 20.0 mL of NaOH.

The base will react with the acid:

NaOH (aq) + HCl (aq)  H2O (l) + NaCl (aq)

NaOH: 0.100 M x 0.0200 L = 0.00200 mol

HCl: 0.100 M x 0.0750 L = 0.00750 mol

NaOH (aq) + HCl (aq)  H2O (l) + NaCl (aq)

0.00200 0.00750 0 0

-0.00200 - 0.00200 +0.00200 + 0.00200

= 0 = 0.00550 = 0.00200 + 0.00200

We have HCl in the flask

pH = -log(0.00550 mol/0.0950L) = -log(0.0580) = 1.237

Determine the pH during the titration of 75.0 mL of 0.100 MHCl by 0.100 MNaOH…

(c) at the equivalence point

The point at which all of the acid is consumed by the added base:

NaOH (aq) + HCl (aq)  H2O (l) + NaCl (aq)

HCl: 0.100 M x 0.0750 L = 0.00750 mol

NaOH: 0.00750 mol (ie 75.0 mL used)

NaOH (aq) + HCl (aq)  H2O (l) + NaCl (aq)

0.00750 0.00750 0 0

-0.00750 - 0.00750 +0.00750 + 0.00750

= 0 = 0 = 0.00750 = 0.00750

We have H2O and salt (neutral) in the flask

pH = 7.000

Determine the pH during the titration of 75.0 mL of 0.100 MHCl by 0.100 MNaOH…

(d) after adding 100.0 mL of NaOH

After the equivalence point, all HCl is gone and excess NaOH is present:

NaOH (aq) + HCl (aq)  H2O (l) + NaCl (aq)

HCl: 0.100 M x 0.0750 L = 0.00750 mol

NaOH: 0.100 M x 0.1000 L = 0.0100 mol

NaOH (aq) + HCl (aq)  H2O (l) + NaCl (aq)

0.0100 0.00750 0 0

-0.00750 - 0.00750 +0.00750 + 0.00750

= 0.0025 = 0 = 0.00750 = 0.00750

We have NaOH in the flask

pOH = -log(0.0025 mol / 0.1750 L) = -log(0.01429) = 1.845

pH = 14.000 – 1.845 = 12.16

(a) Determine the pH during the titration of 75.0 mL of 0.100 M benzoic acid by 0.100 M NaOH before the addition of any NaOH.

Only benzoic acid is present – a weak acid

6.3 x 10−5 ≈ x2 / 0.100 x = 0.00251 approximation is valid (2.5%)

pH = − log 0.00251 = 2.60

(b) Determine the pH during the titration of 75.0 mL of 0.100 M benzoic acid by 0.100 M NaOH after adding 20.0 mL of NaOH.

First consider the stoichiometry of the acid-base reaction

C6H5COOH: 0.100 M x 0.0750 L = 0.00750 mol

NaOH: 0.100 M x 0.0200 L = 0.00200 mol

Benzoic acid and its salt are present – a buffer

C6H5COOH: 0.00550 mol / 0.0950 L = 0.05789 M

C6H5COONa: 0.00200 mol / 0.0950 L = 0.02105 M

Benzoic acid and its salt are present – a buffer 0.100 M benzoic acid by 0.100 M

C6H5COOH: 0.00550 mol / 0.0950 L = 0.05789 M

C6H5COONa: 0.00200 mol / 0.0950 L = 0.02105 M

6.3 x 10−5 ≈ (x)(0.02105) / 0.05789 x = 1.7326 x 10−4 approximation is valid (0.3%)

pH = − log 1.7326 x 10−4 = 3.76

or pH = pKa + log [base]/[acid] = 4.201 + log (0.02105 / 0.05789) = 3.76

(c) Determine the pH during the titration of 75.0 mL of 0.100 M benzoic acid by 0.100 M NaOH at the half-equivalence point (the titration midpoint).

Half-equivalence point is when 37.5 mL NaOH is added to the solution

C6H5COOH: 0.100 M x 0.0750 L = 0.00750 mol

NaOH: 0.100 M x 0.0375 L = 0.00375 mol

Benzoic acid and its salt are present – equal amount – special point

C6H5COOH: 0.00375 mol / 0.1125 L = 0.0333 M

C6H5COONa: 0.00375 mol / 0.1125 L = 0.0333 M

Benzoic acid and its salt are present – a buffer 0.100 M benzoic acid by 0.100 M

C6H5COOH: 0.00375 mol / 0.1125 L = 0.0333 M

C6H5COONa: 0.00375 mol / 0.1125 L = 0.0333 M

6.3 x 10−5 ≈ (x)(0.0333) / 0.0333 x = 6.3 x 10−5

pH = − log 6.3 x 10−5= 4.20

or pH = pKa + log [base]/[acid] = 4.201 + log (0.0333 / 0.0333) = 4.20

or at half-equivalence pH = pKa (for weak-strong acid-base scenario)

(d) Determine the pH during the titration of 75.0 mL of 0.100 M benzoic acid by 0.100 M NaOH at the equivalence point.

Equivalence point is when 75.0 mL NaOH is added, all acid and base is consumed

C6H5COOH: 0.100 M x 0.0750 L = 0.00750 mol

NaOH: 0.100 M x 0.0750 L = 0.00750 mol

only the salt is present – basic solution

C6H5COONa: 0.00750 mol / 0.1500 L = 0.0500 M

only the salt is present – basic solution 0.100 M benzoic acid by 0.100 M

C6H5COONa: 0.00750 mol / 0.1500 L = 0.0500 M

Kb = 1.59 x 10−10 ≈ x2 / 0.0500 x = 2.82 x 10−6

pOH = − log 2.82 x 10−6 = 5.550

pH = 8.45

(e) Determine the pH during the titration of 75.0 mL of 0.100 M benzoic acid by 0.100 M NaOH after adding 100.0 mL of NaOH.

The stoichiometry of the acid-base reaction

C6H5COOH: 0.100 M x 0.0750 L = 0.00750 mol

NaOH: 0.100 M x 0.1000 L = 0.0100 mol

NaOH and the salt are present – NaOH is a strong base

NaOH: 0.00250 mol / 0.1750 L = 0.0143 M

pOH = 1.845; pH = 12.155

25. Determine 0.100 M benzoic acid by 0.100 M the pH during the titration of 75.0 mL of 0.100 M ammonia (Kb = 1.8  10–5) by 0.100 MHCl at the following points.

(a) Before the addition of any HCl11.13

(b) After adding 20.0 mL of HCl9.69

(c) At the titration midpoint 9.26

(d) At the equivalence point 5.28

(e) After adding 100.0 mL of HCl1.845