Consider the reaction IO−4(aq)+2H2O(l)⇌H4IO−6(aq);Kc=3.5×10−2IO4−(aq)+2H2O(l)⇌H4IO6−(aq);Kc=3.5×10−2 If you start with 22.0 mLmL of a 0.908 MM solution of...

Consider the reaction IO−4(aq)+2H2O(l)⇌H4IO−6(aq);Kc=3.5×10−2IO4−(aq)+2H2O(l)⇌H4IO6−(aq);Kc=3.5×10−2
If you start with 22.0 mLmL of a 0.908 MM solution of NaIO4NaIO4, and then dilute it with water to 500.0 mLmL, what is the concentration of H4IO−6H4IO6− at equilibrium?

Expert Solution

First, concentration of final solution of NaIO4 will be found. Then, using ice charts, concentration of H4IO6- will be found out.

queen_honey_blossom answered 7 hours ago

Consider the reaction IO−4(aq)+2H2O(l)⇌H4IO−6(aq);Kc=3.5×10−2 If you start with 25.0 mL of a 0.905 M solution of...

Consider the reaction IO−4(aq)+2H2O(l)⇌H4IO−6(aq);Kc=3.5×10−2 If you start with 25.0 mL of a 0.905 M solution of NaIO4, and then dilute it with water to 500.0 mL, what is the concentration of H4IO−6 at equilibrium?

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