Respuesta :
Answer : The concentration in (M)of bromide ions in a saturated solution of mercury (II) bromide is, [tex]2.7\times 10^{-7}M[/tex]
Explanation :
The solubility equilibrium reaction will be:
[tex]HgBr_2\rightleftharpoons Hg^{2+}+2Br^{-}[/tex]
Let the molar solubility be 's'.
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Hg^{2+}][Br^{-}]^2[/tex]
[tex]K_{sp}=(s)\times (2s)^2[/tex]
[tex]K_{sp}=4s^3[/tex]
Given:
[tex]K_{sp}[/tex] = [tex]8.0\times 10^{-20}[/tex]
Now put all the given values in the above expression, we get:
[tex]K_{sp}=4s^3[/tex]
[tex]8.0\times 10^{-20}=4s^3[/tex]
[tex]s=2.7\times 10^{-7}M[/tex]
Therefore, the concentration in (M)of bromide ions in a saturated solution of mercury (II) bromide is, [tex]2.7\times 10^{-7}M[/tex]
