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Tema [17]
3 years ago
12

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.
Chemistry
1 answer:
rosijanka [135]3 years ago
3 0

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}

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