Question

A stock solution of sodium salicylate 5% in water is prepared. Thirty mL of the stock solution is titrated with 0.1N HCl until precipitation occurs. What will be the pH of the solution at the point of permanent precipitate formation? (Solubility of salicylic acid at 25 C is 1 gm in 460 ml water; M.W. of Salicylic acid is 138.1; M.W. of sodium salicylate is 160.1; pKa =2.97).

Answer 1

The concentration of 5% sodium salicylate solution in water

=(5*1000)/(160.1*100) N.=0.31 N .

Therefore required 0.1 N HCl for precipitation is

S₁V₁=S₂V₂ ,V₁=S₂V₂/S_1,

V1=0.31*30/.1 ml=93ml

Therefore 93ml 0.1N HCl is required for precipitation= 93 ml salicylic acid solution.

1 gm salicylic acid dissolved in 460 ml of water.

Therefore 93 ml of solution contains =93/460 g=0.202g

Therefore the concentration of 0.202g salicylic acid =(0.202*1000)/(138.1*93)

=0.015 N.

Therefore P^H=P^Ka+log[0.015]/[0.1]

P^H=(2.97-0.82)=2.15