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4 years ago

The molar solubility if Pb(IO3)2 in

Chemistry

1 answer:

lana66690 [7]4 years ago

7 0

Answer:

Ksp = 2.4 * 10^-13

Explanation:

Step 1: Data given

Molarity of NaIO3 = 0.10 M

The molar solubility of Pb(IO3)2 = 2.4 * 10^-11 mol/L

Step 2: The initial concentration

NaIO3 = 0.1M

Na+ = 0 M

2IO3- = 0 M

Step 3: The concentration at the equilibrium

All of the NaIO3 will react (0.1M)

At the equilibrium the concentration of NaIO3 = 0 M

The mol ratio is 1:1:1

The concentration of Na+ and IO3- is 0.1 M

Pb(IO3)2 → Pb^2+ + 2IO3^-

The concentration of Pb(IO3)2 can be written as X

The concentration of Pb^2+ can be written as X

The concentration of 2IO3^- can be written as 2X

Ksp = (Pb^2+)(IO3^-)²

⇒ with (Pb^2+) = 2.4*10^-11

⇒ with (IO3^-) = 2x from the Pb(IO3)2 and 0.1M from the NaIO3.

⇒The total (IO3^-) = 2x + 0.1 and we assume that x is so small that we can neglect it.

Ksp = (2.4 *10^-11)*(0.1)²

Ksp = 2.4 * 10^-13

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