Question

Let us write the appropriate equilibria and associate the correction le="K_b" alt="K_b" align="absmiddle" class="latex-formula"> 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 $K_a_1$ = 1.4 × 10⁻² and $K_a_2$ = 6.3 × 10⁻⁸.

Answer 1

Explanation:

The relation between $K_a\&K_b$ is given by :

$K_w=K_a\times K_b$

Where :

$K_w=1\times 10^{-14}$ = Ionic prodcut of water

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

The value of $K_{b1}$:

$1\times 10^{-14}=1.4\times 10^{-2}\times K_{b1}$

$K_{b1}=\frac{1\times 10^{-14}}{1.4\times 10^{-2}}=7.1\times 10^{-13}$

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

The value of $K_{b2}$:

$1\times 10^{-14}=6.3\times 10^{-8}\times K_{b1}$

$K_{b1}=\frac{1\times 10^{-14}}{6.3\times 10^{-8}}=1.6\times 10^{-7}$