Question

Which of the following aqueous solutions would be the best to use for selective removal of the Hg22+ ions from a solution containing both Hg22+ ions and Pb2+ ions? (hint: consult a table of solubility rules, or the table of Ksp values on page A-13 of the 6th edition Silberberg "Chemistry" textbook, or page 828 of the Silberberg "Principles of General Chemistry" 2nd Edition textbook, or some other suitable source of such information)

NaF(aq)

Na2SO4(aq)

NaI(aq)

NaNO3(aq)

Answer 1

Sodium iodide solution will be the best to use , as the solubility product value of Hg₂I₂ very small, 4.5 x 10^-29, which means it can be precipitated even with low concentrtion of iodide ions, while the ksp of lead iodie is quite high ,8.5x 10 ^-9, means to precipitate lead we need much higher concentration of iodide.

Thus iodide will be ideal to remove all Hg₂⁺² ions from lead solution.

adding sodium sulphide can precipitate both mercurous and lead ions as the ksp values of their sulphates are close, 6.8x 10^-7 and 1.8 x10^-8 respectively.

In general all the nitrates are soluble in water, so sodium nitrate is not an option.