Final answer:
To calculate the ion concentrations of Mercury(I) bromide from the Ksp, establish the Ksp expression, define the molar solubility as 's', and solve for 's'. The concentration of Hg2^2+ would be 's' and for Br- would be 2s.
Step-by-step explanation:
Calculating Ion Concentration from Ksp
To estimate the concentration of mercury ions and bromide ions for Mercury(I) bromide (Hg2Br2), we start by writing the dissociation equation and the expression for the solubility product constant (Ksp).
Hg2Br2(s) ⇌ Hg22+(aq) + 2 Br-(aq)
The Ksp expression for Mercury(I) bromide is then written as:
Ksp = [Hg22+][Br-]2
Given the Ksp is 5.6 x 10-23, if we let the molar solubility of Hg2Br2 be 's,' the equilibrium concentrations will be [Hg22+] = s and [Br-] = 2s because for every mole of Hg2Br2 that dissolves, one mole of Hg22+ ions and two moles of Br- ions are produced.
Plugging these into the Ksp expression yields:
5.6 x 10-23 = (s)(2s)2
5.6 x 10-23 = 4s3
Solving for 's' we get:
s = ∛(5.6 x 10-23 / 4)
The value of 's' is the concentration of Hg22+ ions, and 2s is the concentration of Br- ions. This method allows us to estimate ion concentrations in a saturated solution given the solubility product constant.