Consider the reaction io−4(aq)+2h2o(l)⇌h4io−6(aq); kc=3.5×10−2 if you start with 26.0 ml of a 0.904 m solution of naio4, and then dilute it with water to 500.0 ml, what is the concentration of h4io−6 at equilibrium?

0.744 M

Explanation:

IO⁻⁴(aq) + 2H₂O(l) ⇌ H₄IO⁻⁶(aq)

Kc = 3.5×10⁻²= [H₄IO⁻⁶] / [IO⁻⁴]

First let's calculate the new concentration of IO⁻⁴ at equilibrium:

0.904 M * 26.0 mL / 500.0 mL = 0.047 M = [IO⁻⁴]

Now we can calculate [H₄IO⁻⁶] using the formula for Kc:

3.5×10⁻²= [H₄IO⁻⁶] / [IO⁻⁴]

3.5×10⁻²= [H₄IO⁻⁶] / 0.047 M

[H₄IO⁻⁶] = 0.744 M

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.