Question

The potassium ion in 250mL of a water sample was precipitated with tetraphenylborate. The precipitate was filtered and dissolved in an organic solvent. The solution was treated with Hg (II) -EDTA following the reaction:

  HgY2- + (C6H5)4B- + 4H2O = H3BO3 + 4C6H5Hg+ + 4HY3- + OH-

The liberated EDTA was titrated with 28.73 mL of Zn2 + 0.0437M. Find the concentration of K + in the original sample.

Answer 1

Potassium ion is precipitated with tetraphenylborate is:

K⁺(aq) + B(C₆H₅)₄^-(aq) KB(C₆H₅)₄ (s)

This precipitates are then washed, filtered and redissolved in organic solvent which rects with Hg(II) -EDTA to form potassium - EDTA complex. It reacts with Zn²⁺ to form Zn-EDTA complex.

Reaction is:

4HgY^2- + B(C₆H₅)₄^- + 4H₂O H₃BO₃ + 4C₆H₅Hg⁺ + 4HY^3- + OH^-

From reaction, 1 mol of Hg(II)- EDTA produces 1 mol of EDTA liberated

Also, 1 mol of EDTA liberated reacts with 1 mol of Zn²⁺

Molarity of Zn²⁺ = 0.0437 M = 0.0437 mmol/mL

Volume of  Zn²⁺ = 28.73 mL

mmol of  Zn²⁺ = Molarity * Volume in mL = 0.0437 mmol/mL* 28.73mL = 1.2555 mmol  Zn²⁺ = mmol of potassium - EDTA complex

Now, 4 mol of K⁺ reacts with 1 mol of EDTA to form potassium - EDTA complex

Thus, mmol of K⁺ = mmol of EDTA/4 = 1.2555 mmol /2 = 0.31387 mmol K⁺

Molar mass of potassium = 39.098 g/mol = 39.098 mg/mmol

Thus, Mass of potassium in sample = mmoles * molar mass =0.31387 mmol*39.098 mg/mmol = 12.2719 mg

Volume of solution = 250 mL = 0.250 L

concentration of K⁺ in ppm = mass of K⁺ / Volume = 12.2719 mg/ 0.250 L = 49.088 mg/L =49.088 ppm