Let us write the appropriate equilibria and associate the correction [tex]K_b[/tex] values. Remember, we will want to calculate the concentrations of all species in a 0.390 M Na ₂SO₃ (sodium sulfite) solution. The ionization constants for sulfurous acid are [tex]K_a_1[/tex] = 1.4 × 10⁻² and [tex]K_a_2[/tex] = 6.3 × 10⁻⁸.

Respuesta :

Explanation:

The relation between [tex]K_a\&K_b[/tex] is given by :

[tex]K_w=K_a\times K_b[/tex]

Where :

[tex]K_w=1\times 10^{-14}[/tex] = Ionic prodcut of water

The value of the first ionization constant of sodium sulfite = [tex]K_{a1}=1.4\times 10^{-2}[/tex]

The value of [tex]K_{b1}[/tex]:

[tex]1\times 10^{-14}=1.4\times 10^{-2}\times K_{b1}[/tex]

[tex]K_{b1}=\frac{1\times 10^{-14}}{1.4\times 10^{-2}}=7.1\times 10^{-13}[/tex]

The value of the second ionization constant of sodium sulfite = [tex]K_{a2}=6.3\times 10^{-8}[/tex]

The value of [tex]K_{b2}[/tex]:

[tex]1\times 10^{-14}=6.3\times 10^{-8}\times K_{b1}[/tex]

[tex]K_{b1}=\frac{1\times 10^{-14}}{6.3\times 10^{-8}}=1.6\times 10^{-7}[/tex]