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

`pH=6.07 .`

Step-by-step explanation:

It is given that pH of
`H_2CO_3=4.3* 10^(-7)`.

Now, molality of
`NaHCO_3` , M=
`(No\ of\ moles)/(Volume\ in\ liters)=(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* 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.