Question

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?

Answer 1

Answer:

$pH=6.07 .$

Explanation:

It is given that pH of $H_2CO_3=4.3\times 10^{-7}$.

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

Now, we know,

        $pH=pK_a-log\ C.$   ...1

Here C is concentration.

Now, $pK_a=-logK_a=-log(4.3\times 10^-7)=6.37 .$

Putting all values in equation 1.

We get,

$pH=6.37-0.30=6.07 .$

Hence, this is the required solution.