Question

A 800.0mg sample of solid lead(II) iodide (Ksp = 9.8*10^-9) is dissolved in 5.0L of pure water (at 25 C). How much hydroiodic acid, a strong acid, must be added to the solution (in g to two decimal places) to begin precipitating lead(II) iodide out of solution? (ignore volume changes by the addition solutes) A: 2.95g

Answer 1

mass of PbI2 = 800mg   = 0.800 g

moles of PbI2 = mass/molar mass

                  = 0.800/461

                 = 0.0017 mol

moles of I- = 2 x 0.0017 = 0.0034 --------------------> 1

Volume of pure water = 5.0 L

molarity of PbI2 = moles/volume

                     = 0.0017/ 5.0

                     = 0.00035 M

Ksp = 9.80 x 10^{^-9}

Ksp = [Pb2+] [I-]^{^2}

9.80 x 10^-9 = 0.00035 [I-]^{^2}

[I-] = 0.0053 M

Moles of I- = 0.0053 x 5.0

             = 0.0265 ------------------------>2

now do ( 2 )- (1)

moles of HI needed to added = 0.0265-0.0034

                                                = 0.0231

mass of HI = moles x molar mass

                  = 0.0231 x 128

                = 2.95 g