Question #67a25

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Answer 1 · 2017-03-20T04:44:06

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

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#