Question

(a) Saturated Sodium Chloride. Watch Video-3a where to 4 mL of saturated (5.4 M) NaCl in a 13 x 100mm test tube, 2 mL of concentrated (12 M) HC) is added. Write down the chemical reaction(s) involved, being sure to include the state of matter of all reactants and products so it is clear if there is a solid/precipitate expected. (1 point) Calculate the total Cl concentration in the NaCl solution after adding 2 mL of 12 M HCl to 4 mL of 5.4 M NaCl. (1 point) Explain what you observed in terms of the saturated solution equilibrium. (1 point)

Answer 1

Answer :

This process is known as salting out process.

A. The reaction equation for this process is given in image.

Reaction equation: Not Low + U u eaq? + Heago tu el paq) t Hilog) + (1 100) Naces) t

B. Cl- ion concentration is also calculated and represented in the image.

To Fal cu comentrations- + Nache) No tit lag) caq). Hos Htcoas + Chian) Awording to this equation 1 molas 7. Nall Hel m gives 1 mudar g co con. Thus 12 M Hu contains 12 M H G 12Ma ion. And 54M Nal gives B4 M Nat Ç Shri ton. Tofw [ujion - 12754. ľuja 17.4M

C. At the saturation point, the dissolution of NaCl is in the equilibrium, which is represented as,

NaCl_(s) == Na⁺_(aq) + Cl^-_(aq)

According to the Le-chatleir principle, if the concentration of any one of the compound in the equilibrium increases, the equilibrium will tend to shift in the opposite direction to equalise the added concentration.

Thus if add con. HCl, the concentration of Cl^- ion increases and the equilibrium will shift backwards to precipitate the NaCl.

Thus white colour solid is observed in the test tube when con. HCl is added to the saturated NaCl solution. This this process is known as salting out process.