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The Ka of carbonic acid (H2CO3) is 4.3 x 10–7. A solution of sodium hydrogen carbonate (NaHCO3) solution is created by dissolving 4.00 moles of sodium hydrogen carbonate in 2.00 L of aqueous solution. What is the pH of the solution at equilibrium?

Respuesta :

Answer:

[tex]pH=6.07 .[/tex]

Explanation:

It is given that pH of [tex]H_2CO_3=4.3\times 10^{-7}[/tex].

Now, molality of [tex]NaHCO_3[/tex] , M=[tex]\dfrac{No\ of\ moles}{Volume\ in\ liters}=\dfrac{4}{2}=2\ molar.[/tex]

Now, we know,

        [tex]pH=pK_a-log\ C.[/tex]   ...1

Here C is concentration.

Now, [tex]pK_a=-logK_a=-log(4.3\times 10^-7)=6.37 .[/tex]

Putting all values in equation 1.

We get,

[tex]pH=6.37-0.30=6.07 .[/tex]

Hence, this is the required solution.

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