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Quantitative analysis of Cl^- ion is often performed by a titration with silver

ID: 529059 • Letter: Q

Question

Quantitative analysis of Cl^- ion is often performed by a titration with silver nitrate, using sodium chromate as an indicator. As standardized AgNQ_3 is added, both white AgCl and red Ag_2CrO_4 precipitate, but so long as some Cl^- remains, the Ag_2CrO_4 redissolves as the mixture is stirred. When the red color is permanent, the equivalence point has been reached. (b) If 32.00 cm^3 of 0.5700 M NaCl is mixed with 32.00 cm^3 of 0.5700 M AgNO_3, what is the concentration of Ag remaining in solution? Is this sufficient to precipitate any silver chromate? 1.4 times 10^-2 M yes (Assume 0.02 M CrO_4^2-) no Enter your answer in scientific notation.

Explanation / Answer

Since the conc and molarity of both NaCl and AgNO3 solutions are the same, so the moles of both the solutions are also same.

This means all the Cl- ions react with all the Ag+ ions in solution. So after the AgCl ppt is formed there are no more Cl- ions or Ag+ ions left in the solution.

After the formation of AgCl ppt, the following dissolution reaction takes place.

AgCl(s) ----> Ag+(aq) + Cl-(aq)

Ksp for AgCl = 1.7*10-10

So due to the dissolution reaction, the Ag+ conc in solution is: (1.7*10-10)0.5 = 1.303*10-5

Ksp for silver chromate is 1.1*10-12

Ionic product for Ag2CrO4 = [Ag+]2*[CrO42-] = (1.303*10-5)*(0.02) = 2.6*10-7

SInce this product is greater than the Ksp for AgCrO4, thus it is sufficient to ppt out AgCrO4.

So answer is YES