Question

A solution contains Ag+ and Hg2+ ions. The addition of 0.100 L of 1.10M NaI solution is just enough to precipitate all the ions as AgI and HgI2. The total mass of the precipitate is 25.5 g. Find the mass of AgI in the precipitate.

Answer 1

mass balance:

Mass of AgI + mass of HGI2 = 25.5

mol balance:

mol of NaOH = MV = 1.1*0.1 = 0.11

mol of I- from AgI + mol of I- from HgI2 = 0.11

mol of AgI + 2*mol of HgI2 = 0.11

then, assume

"x" = mass of AgI

then, mass of HgI2 = (25.5-x)

MW of HgI2 = 454.4

MW of AgI = 234.77

apply mol balance:

mol of AgI + 2*mol of HgI2 = 0.11

mass of AgI/ MW of AgI = mol of AgI

mass of HgI2 / MW of HgI2 = mol of HgI2

mol of AgI + 2*mol of HgI2 = 0.11

mass of AgI/ MW of AgI + 2 * mass of HgI2 / MW of HgI2 = 0.11

substitute

x/ 234.77 + 2 * (25.5-x) / 454.4 = 0.11

solve for x

454.4 x + 2 * (25.5-x) *234.77 = 0.11*234.77 *454.4

454.4 x + 2 * (25.5-x) *234.77  = 11734.74

454.4x - 2*234.77 x + 2*25.5*234.77= 11734.74

x(454.4 + -2*234.77 ) = 11734.74-2*25.5*234.77

x = ( 11734.74-2*25.5*234.77)/(454.4 + -2*234.77 )

x = 15.754 g of AgI