If 0.25 mol of HClO2 is added to enough water to prepare a 0.25 M HClO2 aqueous solution, what is the concentration of H3O+(aq) at equilibrium? At this temperature, Kc is 0.15 for the equation, HClO2(aq) + H2O(l) ⇌ H3O+(aq) + ClO2−(aq)

The concentration of H3O+= 0.15M

Explanation:

From The equation of reaction

HClO2(aq) + H2O(l) ⇌ H3O+(aq) + ClO2−(aq)

0.25mol HClO2(aq) 0.25mol producesClO2−(aq) and x-mol of H3O+

Using Kc = [H3O+][ClO2-]/[HClO2]

0.15= 0.25*x/0.25

Simplify

x=0.15M

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

what is the question?