Question

Determination of the Solubility Product Constant of an Organic Salt Prepurod by Judith C.Foster, Bosndoin College PURPOSE OFTHE EXPERIMENT Determine the solubility and solubility product constant of potassium hydrogen tartrate by titrimetry BACKGROUND INFORMATION A complete characterization of a chemical system involving substances dissolved in water includes studying the relevant equilibria. For the gene ralized chemical reaction aA +bBcC+ dD, the equilibrium constant expression is Eq.1 In the case of an acid salt, such as sodium hydrogen oxalate monohydrate (NaHC,0,-H,O) dissolving in water, an equilibrium is established among solid sodium hydrogen oxalate (NaHC O), sodium ion (Na"), and hydrogen oxalate ion (HCO,-), as shown in Equation 2. Eq. 2) A quantitative characterization of this equilibrium would involve determi- nation of the amount of dissolved NaHC Oslaq) in the presence of excess solid NaHC O4(s). From Equation 2, we see that one mole of Na" and one mole of HC,O, would be present in solution if one mole of NaHC,O dissolved. The molar ratio of dissociated NaHC20, to Na" in solution to HC,O, in solution would be 1:1:1. Thus, by determining the concentration of HC O, in solution, we can find the amount of dissolved NaHC 0. The concentration of HC,O in solution could be determined by the direct titration of the filtered solution with a standard NaOH solution, as Nane Pre-Laboratory Assignment 1. In an appropriate source, read a discussion of the laboratory techniques used in acid-base titrations 2. An experiment similar to the one in this module 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: CoHO,COOH(s) C,H,0,Coo (aq)+H (aq) Titrations using 1.14x 10 'M NaoH were performed, and the following data were obtained determination temperature of solution, "C volume of acid solution titrated, ml volume of NaOH solution used, ml 20.2 25.00 4.61 20.4 22.00 13.02 20.1 20.10 11.73 Calculate the following: number of moles of NaOH used number of moles of H' titrated [H 1 in acid solution ICHsOCOO.1 in acid solution Kop average Ksp solubility of acid in g per 100 ml chemistry handbook value for 1.15 g per 100 mL solubility at 20"C

Answer 1

(1) Given,

Molarity of NaOH (M) = [NaOH] = 1.14 x 10^-1 M

Volume of NaOH solution used (mL)

V₁ = 14.61 mL

Number of moles of NaOH used (n) = Molarity of NaOH (M) x Volume of NaOH solution used (V)

n = MxV

n₁ = MxV₁

n₁ = 1.14 x 10^-1 M x 14.61 mL = 1.66 mmol = 1.66 x 10^-3 mol [Note: 1 M mL = 1 mmol]

Similarly for V₂ = 13.02 mL and V3 = 11.73 mL

n₂ = 1.14 x 10^-1 M x 13.02 mL = 1.48 mmol = 1.48 x 10^-3 mol

n₃ = 1.14 x 10^-1 M x 11.73 mL = 1.34 mmol = 1.34 x 10^-3 mol

(2) We know that the number of moles NaOH used is equal to the number of moles of H⁺ titrated.

n_OH^- = n_H⁺

Hence,

number of moles of H⁺ titrated

n_H⁺(1) = n₁ = 1.66 x 10^-3 mol

n_H⁺(2) = n₂ = 1.48 x 10^-3 mol

n_H⁺(2) = n₃ = 1.34 x 10^-3 mol

(3) [H⁺] in acid solution

We know that the [H⁺] in acid solution is calculated by number of moles of H⁺ titrated divided by volume of acid solution titrated.

[H⁺] = (n_H⁺/V_{acid solution})x1000

[H⁺](1) = (1.66 x 10^-3 mol/25 mL)x1000 = 0.0664 M

Similarly,

[H⁺](2) = (1.48 x 10^-3 mol/22 mL)x1000 = 0.0673 M

[H⁺](3) = (1.34 x 10^-3 mol/20.10 mL)x1000 = 0.0667 M

(4) [C₆H₅O₃COO^-] in acid solution

We know that in the solution concentration of H⁺ is equal to the concentration of C₆H₅O₃COO^-.

[H⁺] = [C₆H₅O₃COO^-]

Hence,

[C₆H₅O₃COO^-](1) = [H⁺](1) = 0.0664 M

[C₆H₅O₃COO^-](2) = [H⁺](2) = 0.0673 M

[C₆H₅O₃COO^-](3) = [H⁺](3) = 0.0667 M

(5) Ksp = [C₆H₅O₃COO^-][H⁺]

Ksp (1) = 0.0664 M x 0.0664 M = 4.41 x 10^-3

Ksp (2) = 0.0673 M x 0.0673 M = 4.53 x 10^-3

Ksp (3) = 0.0667 M x 0.0667 M = 4.45 x 10^-3

(6) average Ksp = [Ksp (1) + Ksp (2) + Ksp (3)]/3

average Ksp = [4.41 x 10^-3 + 4.53 x 10^-3 + 4.45 x 10^-3]/3 = 4.46 x 10^-3

(7) solubility of acid (S)= (Ksp)^1/2 = (4.46 x 10^-3) = 0.0668 M

S in g/L = 0.0668 M x Molar Mass of acid = 0.0668 M x 170.12 g/mol = 11.36 g/L

S in g per 100 mL = 11.36 g/ 1000 mL = 1.136 g/ 100 mL

S in g per 100 mL = 1.14