Question #133448

Calculate the molar solubility of Pb (IO3)2 in 1.0 x 10–4 M Pb (NO3)2. Ksp =2.5 x10-13

Expert's answer

In the solution of 1.0x10^{-4} M Pb(NO_{3})_{2 }the concentrations of ions in solution is thus;

[Pb^{2+}] = 1.0 x 10^{-4}M

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

On addition of Pb(IO_{3})_{2}, the solution will dissociate partly as;

Pb(IO_{3})_{2 } Pb^{2+}_{(aq)} + 2IO_{3}^{- }_{(aq)}

_{ } ** I ** = 1.0x10^{-4 }0

** C** = X 2X

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

Therefore; K_{sp }

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;

Since, [2X]^{2} = 4X^{2 }

and (1.0x10^{-4} )(4X^{2}) = 4.0x10^{-4}X^{2}

Therefore, 2.5x10^{-13} = 4.0 x10^{-4}X^{2}

^{ }Dividing both sides by 4.0 x10^{-4 }gives,

6.25x10^{-10} = X^{2}

Removing the exponent

(6.25x10^{-10}) = X^{2}

2.5x10^{-5} = X

Therefore the molar solubility of P(IO_{3})_{2} in 1.0 x 10^{–4} M Pb (NO_{3})_{2 }is **2.5x10**^{-5}** M**

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