The solubility of lead (II) Iodate, #Pb(IO_3)_2#, is 0.76 g/L at 25*C. How do you calculate the Value of Ksp at this same temperature?

Answer 1

#K_(sp)# #Pb(IO_3)_2# #=# #??#

We can represent the solubility of #Pb(IO_3)_2# as:
#Pb(IO_3)_2(s) rightleftharpoons Pb^(2+) + 2IO_3^-#
#K_(sp)# #=# #[Pb^(2+)][IO_3^-]^2#
And if we let #S="solubility of lead iodate"#, then,
#K_(sp)# #=# #(S)(2S)^2# #=# #4S^3#.
So now we work out the solubility of #Pb(IO_3)_2#.
We are given #S_("mass")=0.76*g*L^-1#
#S_("molar")=(0.76*g)/(557.00*g*mol^-1)xx1/(1*L)#
#=# #1.37xx10^-3*mol*L^-1#
And thus #K_(sp)=4xx(1.37xx10^-3)^3# #=# #??#

Lead iodate is thus quite an insoluble beast.

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Answer 2

[ \text{K}_{\text{sp}} = [\text{Pb}^{2+}][\text{IO}_3^-]^2 ]

Given the solubility of lead (II) iodate ((\text{Pb(IO}_3)_2)) is 0.76 g/L, convert this to moles per liter. Use the stoichiometric coefficients to determine the concentration of lead ions ((\text{Pb}^{2+})) and iodate ions ((\text{IO}_3^-)). Substitute these values into the Ksp expression to find the Ksp value at 25°C.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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