An experiment was performed to determine the solubility and the solubility produ

Question

An experiment was performed to determine the solubility and the solubility product constant of gallic acid, a monoprotic organic acid (gmm=170.12). The equilibrium involved is: C6H5O3COOH(s) <-----> C6H5O3COO-(aq) + H+(aq)

Titrations using 1.14x10^-1M NaOH were performed.

Determination 1 2 3

Temp of solution, C 20.2 20.4 20.1

Vol of acid solution titrated, mL 25.00 22.00 20.10

Vol of NaOH solutions used, mL 14.61 13.02 11.73

Calculate the following:

1. number of moles of NaOH used

2. number of moles of H+ titrated

3. [H+] in acid solution

4. [C6H5O3COO-] in acid solution

5. Ksp

6. Solubility of acid in g per 100mL

Please solve all of these questions for each determination and show work!! Thank you so much.

Explanation / Answer

1.

Amount of 0.114M NaOH volume used = 14.61+13.02+11.73= 39.36ml

Amount of NaOH moles used = 0.114/1000x39.36 = 0.00449

So 0.00449 moles of NaOH used.

2.

Mole ratio NaOH:H+ =1:1

So amount of H+ used for titration is 0.00449 moles

3.

Total volume of H+ used = 25.00+22.00+20.10= 67.1ml

[H+] = (0.00449/67.1)x1000 = 0.0669M

[H+] = 0.0669M

4.

Mole ratio H+:C6H5O3COO- = 1:1

So [H+] = [C6H5O3COO-] = 0.0669M

[C6H5O3COO-] = 0.0669M

5.

Ksp = [H+][C6H5O3COO-] = 4.48x10^-3M^2

Ksp = 4.48x10^-3M^2

6.

Dissolved [H+] or [C6H5O3COO-] = 0.0669M

Solubility of C6H5O3COOH in 1L = 0.0669

Molar mass of C6H5O3COOH = 170.12g/mol

Solubility of C6H5O3COOH in 1L = 0.0669x170.12 = 11.381g

Solubility of C6H5O3COOH in 100ml = 11.381/10 = 1.138g

So the solubility per 100ml is 1.138g

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