Calculate the molar solubility of AgI in the following. Ksp for AgI is 8.30 ×10−17.(a) pure water×10MEnter your answer in scientific notation.(b) 0.0400 M NaI×10MEnter your answer in scientific notation.

(a) solubility in pure water =9.11 x 10⁻⁹  (M)

(b) solubility in 0.08 (M) NaI = 1.04 x 10⁻¹⁵ (M)

Explanation:

(1)                              

                             AgI (s) + H₂O ⇄ Ag⁺(aq) + I⁻ (aq)

                                                          s             s

Solubility product of AgI

Ksp = [Ag⁺] [I⁻]

⇒Ksp = s²

⇒s

solubility in pure water

(2)      

                             Ksp = [Ag⁺] [I⁻]

                          ⇒ 8.3 X 10⁻¹⁷ = S x (0.08+S)

                          and assuming S is small relative to 0.08, due to common          ion effect (I⁻) solubility of salt decreases

                                                            S + 0.08 = 0.08

                                                 ⇒S X 0.08 = 8.3 X 10⁻¹⁷

                                        ⇒S = 1.04 X 10⁻¹⁵ (M) solubility in 0.08 (M) NaI

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

c, liquids evaporate, creating a gas