Question

How many miligrams of Au3+ ions are
present in 1L saturated solution of gold (III)
jodide ?
(Ksp = 2.7 x 10-47 and atomic mass of Au =
197)​

Answer 1

Answer:

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Explanation:

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Answer 2

There are 5.77$\times 10^{-8}$ milligrams of iron(III) ions.

Explanation:

It is given that the solution is that of gold iodide.

Writing the dissociation equilibrium equation

$AuI_3 \rightleftharpoons Au^{3+}(aq) + 3I^-(aq)$

Let the molar solubility of ions be 'S', then,

Solubility product, K$_{sp}$ $= (S) (3S)^3$

Also, it is given that the solubility product is 2.7$\times 10^{-47}$

Substituting the value, we obtain

$2.7\times 10^{-47}= (S) (3S)^3$

$27 S^4 = 2.7\times 10^{-47}$

$S^4 = \frac{2.7\times 10^{-47}}{27}$

$S^4 = 10^{-48}$

$S = 10^{-12}$

S is the molar solubility that is solubility in terms of moles per liter

Now, we have to convert molar solubility in terms of milligrams per liter.

1 mole of gold iodide contains = 577 g of gold iodide (molar mass of gold iodide)

Solubility in terms of grams/L will be S = $10^{-12}\times 577$

S = $5.77\times10^{-10}$

Dividing by 100, to get in terms of grams

S = $5.77\times10^{-8}$