Question

Calculate the concentration of Hg2+ ions in a saturated solution of HgS(s). How many Hg^2+ ions are in 1000 L of the solution.

Answer 1

The concentration of
`Hg^(2+)` is
`[Hg^(2+)]= 1.26 *10^(-27) mol/L`

The number of
`Hg^(2+)` ion is
`A = 0.7588\ ions`

Step-by-step explanation:

Generally the solubility product constant of HgS(s) is
`k_(sp) = 1.6*10^(-54)`

This solubility product constant is mathematically represented as

`K_(sp) = [Hg^(2+)][S^(2-)]`

Since the HgS is saturated it implies that the concentration of sulfur ion is the same as that of mercury ion s

`[Hg^(2+)]= [S^(2-)] = z`

=>
`K_(sp) =z^2`

=>
`z = \sqrt{K_(sp)}`

substituting values

`z = \sqrt{1.6 *10^(-54)}`

`z = 1.26*10^(-27)`

=>
`[Hg^(2+)]= [S^(2-)] = 1.26 *10^(-27) mol/L`

From above
`1.26 *10^(-27)` mole of
`[Hg^(2+)]` is equal to 1 L

Then x mole of
`[Hg^(2+)]` will be equal to 1000 L

Therefore

`x = (1.26*10^(-27) *1000)/(1)`

=>
`x = 1.26*10^(-24) \ moles`

Now the number of
`Hg^(2+)` ions are in 1000 L of the solution is mathematically represented as

`A = x * (N_A)/(1 mole)`

Where
`N_A` is the avogadro's constant which has a value of

`N_A = 6.022*10^(23) \ ions`

Substituting value

`A = 1.26 *10^(-24) * (6.022 *10^(23))/(1)`

`A = 0.7588\ ions`