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

43.452

step-by-step explanation:

284 times .153 = 43.452

is there any options? maybe im just not seeing them but i need the options to

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