In an appropriate source, read a discussion of the laboratory techniques used in

Question

In an appropriate source, read a discussion of the laboratory techniques used in acid-base titrations. An experiment similar to the one in this module was performed to determine the solubility and the solubility product constant of gallic avid, a monoprotic organic acid (gmm = 170.12). The equilibrium involved is: C_6H_5O_3COOH(s) C_6H_5O_3COO^-(aq) + H^+(aq) Titrations using 1.14 times 10^-1 M NaOH were performed, and the following data were obtained. Calculate the following: number of moles of NaOH used number of moles of H* titrated [H^+] in acid solution (C_6H_5O_3COO^-] in acid solution K_sp average K_sp solubility of acid in g per 100 mL chemistry handbook value for solubility at 20 degree C 1.15 g per 100 mL

Explanation / Answer

Moles of NaOH used

Trial 1:

mol of NaOH = M*V = (1.14*10^-1)(14.61*10^-3) = 0.001665 mol of NaOH

Trial 2:

mol of NaOH = M*V = (1.14*10^-1)(13.02*10^-3) = 0.00148 mol of NaOH

Trial 3:

mol of NaOH = M*V = (1.14*10^-1)(11.73*10^-3) = 0.001337 mol of NaOH

b)

Moles of H+ titrated, by definition

mol of H+ = mol of OH-

since this is a neutralization with ratio 1:1

so

Trial 1:

mol of NaOH = M*V = (1.14*10^-1)(14.61*10^-3) = 0.001665 mol of NaOH

then mol of HCl = 0.001665 mol

Trial 2:

mol of NaOH = M*V = (1.14*10^-1)(13.02*10^-3) = 0.00148 mol of NaOH

then mol of HCl = 0.00148 mol

Trial 3:

mol of NaOH = M*V = (1.14*10^-1)(11.73*10^-3) = 0.001337 mol of NaOH

then mol of HCl = 0.001337 mol

c)

[H+] in acid solution is given by

[H+] = mol of H+ / Volume of acid

so...

Trial 1:

[H+] = mol/V = 0.001665/(25*10^-3) = 0.0666

Trial 2:

[H+] = mol/V = 0.00148/(22*10^-3) = 0.06727

Trial 3:

[H+] = mol/V = 0.001337/(20.10*10^-3) = 0.066517

By definition

[Conjugate acid] = [H+]

so

Trial 1:

[H+] = mol/V = 0.001665/(25*10^-3) = 0.0666

[C6H5O3] = 0.0666 M

Trial 2:

[H+] = mol/V = 0.00148/(22*10^-3) = 0.06727

[C6H5O3] = 0.06727 M

Trial 3:

[H+] = mol/V = 0.001337/(20.10*10^-3) = 0.066517

[C6H5O3] = 0.066517 M

For Ksp:

Ksp = [C6H5O3][H+]

Ksp1 = 0.0666*0.0666 = 0.0044355

Ksp2 = (0.066517)(0.066517) = 0.0044245

Ksp3 = (0.06727)(0.06727) = 0.0045252

Ksp avg = (0.0044355+0.0044245+0.0045252)/3 = 0.00446

Solubility per 100 mL:

S = sqrt(Ksp) = sqrt(0.00446) = 0.066783 M of gallic acid

so

MW acid = 170.12

mass solubility = 170.12*0.066783 = 11.3611 g /L

per 100 mL = 11.3611/0.1 = 1.136 g/L

approx value is 1.15 g per 100 mL so we are pretty near

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