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