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 25.0 mLmL of a 0.909 MM solution of NaIO4NaIO4, and then dilute it with water to 500.0 mLmL, what is the concentration of H4IO−6H4IO6− at equilibrium?
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Ответ:
Answer : The concentration of at equilibrium is, 0.00154 M
Explanation :
First we have to calculate the diluted concentration.
where,
are the initial molarity and volume of .
are the final molarity and volume of diluted .
We are given:
Putting values in above equation, we get:
Now we have to calculate the concentration of at equilibrium.
The given chemical reaction is:
Initial conc. 0.0454 0
At eqm. (0.0454-x) x
The expression for equilibrium constant is:
Now put all the given values in this expression, we get:
x = 0.00154 M
Thus, the concentration of at equilibrium is, 0.00154 M
Ответ: