Question #67a25

1 Answer
Mar 20, 2017

(2.5*10^-11"mol")/L

Explanation:

When Pb(IO)_3 dissociates in water it forms

Pb^+ + IO_3^-

If x = molar solubility then

"Therefore Ksp" = [Pb^+][IO_3^-]^2

"Or Ksp"(2.5*10^-13) = [x][2x]^2

As it is in a NaIO_3 of 0.100M and you want the molar solubility of Pb(IO)_3 in NaIO3 and not in water. .

2.5*10^-13 = [x][x + 0.1M]^2

But x is so so so so small that we can ignore it.
Therefore x + 0.1 = 0.1

Solve for x

2.5 * 10^-13 = [x][0.1M]^2

2.5*10^-13 = 0.01x

x = (2.5*10^-13)/0.01

x = 2.5*10^-11

= (2.5*10^-11"mol")/L