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?

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

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.