Question

Gold(III) chloride, AuCl3, has Ksp = 3.2 x 10-25. Calculate the molar solubility of gold(III) chloride in pure water and in various aqueous solutions.

A) Calculate the molar solubility of gold(III) chloride in pure water.

B) Calculate the molar solubility of gold(III) chloride in 0.010 M HCl solution.

Answer #1

The equilibrium will be

AuCl3 ß> Au3+ + 3Cl-

and the solubility product for the equilibrium is that

Ksp = [ Au3+ ] [ Cl-]^3

For (A)

let us take, n, moles/L AuCl3 is dissolved.

So, we get n molar Au3+ and 3n molar Cl-

Ksp = 3.2*10^-25 = n* (3n)^3 = 27 n^4

n^4 = 3.2*10^-25/27 = 1.185*10^-26

So, n = 3.3*10^-7 M, solubility in pure water.

For (B)

Now let us consider the solubility of AuCl3 = m M

We can consider 1L of 0.010 M HCl

moles of extra [ Cl-] = 0.010 M

Total [ Cl-] = (3m + 0.010) M

Now, Ksp = 3.2*10^-25 = (m)*( 3m+0.010 )^3

Since m is small compared to 0.010, for simplifying the calculation
we can assume 3m+0.010= 0.010

3.02*10^-25 = m* 0.010^3

m = 3.02*10^-25/0.010^3 = 3.02*10^-19 M, solubility in 0.010 M HCl
solution.

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