Answer to Question #133448 in General Chemistry for hi

Question #133448
Calculate the molar solubility of Pb (IO3)2 in 1.0 x 10–4 M Pb (NO3)2. Ksp =2.5 x10-13
Expert's answer

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

[NO3-] = 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 Pb2+(aq) + 2IO3- (aq)

I = 1.0x10-4 0

C = X 2X

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

Therefore; Ksp

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

Thus our equation becomes;

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,

6.25x10-10 = X2

Removing the exponent

(6.25x10-10) = X2

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

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