Question

ph of 0.1M NaHCO3(pka1 and pka2 of H2CO3 are 7 and 11)​

Answer 1

Given:

concentration of NaHCO₃ = 0.1M

pKa₁ of H₂CO₃ = 7

pKa₂ of H₂CO₃ = 11

To find:

pH of NaHCO₃ solution

Solution:

Balance chemical equation of given solution is-

NaHCO₃ + H₂O → H₂CO₃ + NaOH

so,

       $K_b =\frac{K_w}{K_a}$

Here $K_1 = K_a$ = dissociation constant of acid

        $K_b$ = dissociation constant of base

        $K_w$ = dissociation constant of water = 10⁻¹⁴

It is given, pKa₁ = 7

and,      pKa₁ = -log[Ka]

             Ka = antilog(-7)

             Ka = 10⁻⁷

therefore,

                   $K_b = \frac{10^-^1^4}{10^-^7}$

                   $K_b = 10^-^7$

As the dissociation is low, therefore

               $K_b = \frac{[OH^-]}{C}$

By substituting all the values, we get

              $10^-^7 = \frac{[OH^-]}{0.1}$

therefore,       [OH⁻] = 10⁻⁷ × 0.1

                      [OH⁻] = 10⁻⁸

Now, calculate pOH as follows,

    pOH = - log[OH⁻]

    pOH = - log(10⁻⁸)

    pOH = 8

Therefore, pH is calculated as,

       pH + pOH = 14

        pH + 8 = 14

        pH = 14 - 8

       pH = 6

So, the pH of 0.1M NaHCO₃ is 6.