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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?

Respuesta :

Answer : The concentration of [tex]H_4IO_6^-[/tex] at equilibrium is, 0.00154 M

Explanation :

First we have to calculate the diluted concentration.

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume of [tex]NaIO_4[/tex].

[tex]M_2\text{ and }V_2[/tex] are the final molarity and volume of diluted [tex]NaIO_4[/tex].

We are given:

[tex]M_1=0.909M\\V_1=25.0mL\\M_2=?\\V_2=500.0mL[/tex]

Putting values in above equation, we get:

[tex]0.909M\times 25.0mL=M_2\times 500.0mL\\\\M_2=0.0454M[/tex]

Now we have to calculate the concentration of [tex]H_4IO_6^-[/tex] at equilibrium.

The given chemical reaction is:

                     [tex]IO_4^-(aq)+2H_2O(l)\rightleftharpoons H_4IO_6^-(aq)[/tex]

Initial conc.    0.0454                         0

At eqm.       (0.0454-x)                       x

The expression for equilibrium constant is:

[tex]K_c=\frac{[H_4IO_6^-]}{[IO_4^-]}[/tex]

Now put all the given values in this expression, we get:

[tex]3.5\times 10^{-2}=\frac{(x)}{(0.0454-x)}[/tex]

x = 0.00154 M

Thus, the concentration of [tex]H_4IO_6^-[/tex] at equilibrium is, 0.00154 M

The concentration of [tex]H_4IO_6^-[/tex] at equilibrium is 0.00154 M.

We can find the concentration or volume of the concentrated or dilute solution using the equation:

  • [tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1 and V_1[/tex] are the initial molarity and volume of [tex]NaIO_4[/tex].

[tex]M_2 and V_2[/tex] are the final molarity and volume of diluted [tex]NaIO_4[/tex] .

Given:

M₁=0.909M

V₁=25.0mL

V₂=500.0mL

To find:

M₂=?

On substituting the values in the above formula:

[tex]M_1V_1=M_2V_2\\\\M_2=\frac{M_1V_1}{V_2}\\\\ M_2=\frac{0.909*25}{500} \\\\M_2=0.0454M[/tex]

In order to calculate the concentration of [tex]H_4IO_6^-[/tex] at equilibrium:

The given chemical reaction:

[tex]IO^{-4}(aq)+2H_2O(l)[/tex] ⇄ [tex]H_4IO^{-6}(aq)[/tex]

Initial conc.   0.0454                         0

At ⇄             (0.0454-x)                       x

In order to find equilibrium constant:

[tex]k_c=\frac{[H_4IO^{-6}]}{[IO^{-4}]} \\\\3.5*10^{-2}=\frac{x}{0.0454-x} \\\\x=0.00154M[/tex]

Thus, the concentration of [tex]H_4IO_6^-[/tex] at equilibrium is 0.00154 M.

Find more about "Equilibrium constant" here:

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