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

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.