Answer to Question #133448 in General Chemistry for hi

Question

Calculate the molar solubility of Pb (IO3)2 in 1.0 x 10–4 M Pb (NO3)2. Ksp =2.5 x10-13

Answer 1

In the solution of 1.0x10^-4 M Pb(NO₃)₂ the concentrations of ions in solution is thus;

[Pb²⁺] = 1.0 x 10^-4M

[NO^3-] = 2x 1.0 x10^-4 = 2.0 x 10^-4 M

On addition of Pb(IO₃)₂, the solution will dissociate partly as;

Pb(IO₃)₂ $\to$ Pb²⁺_(aq) + 2IO₃^- _(aq)

I = 1.0x10^-4 0

C = X 2X

E = (1.0x10^-4 + X ) 2X

Therefore; K_sp $=$ $[Pb^2+]^1 2[IO3^-]^2$

$2.5x10^-13 = [1.0x10^-4 + X]^1[2X]^2$

The X in the [1.0x10^-4 + X] is so small that it can be neglected/ignored because from its very low K_sp, it is a greatly insoluble salt

Thus our equation becomes;

$2.5x10^-13 = [1.0x10^-4]^1[2X]^2$

Since, [2X]² = 4X²

and (1.0x10^-4 )(4X²) = 4.0x10^-4X²

Therefore, 2.5x10^-13 = 4.0 x10^-4X²

Dividing both sides by 4.0 x10^-4 gives,

$\dfrac{2.5x10^-13}{4.0x10^-4}=\dfrac{4.0x10^-4X^2}{4.0x10^-4}$

6.25x10^-10 = X²

Removing the exponent

(6.25x10^-10)$^\dfrac{1}{2}$ = X²$^\dfrac{1}{2}$

2.5x10^-5 = X

Therefore the molar solubility of P(IO₃)₂ in 1.0 x 10^–4 M Pb (NO₃)₂ is 2.5x10^-5 M