3 years ago

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Chemistry

1 answer:

givi [52]3 years ago

8 0

Ummm I just need to answer questions sorry!!!

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Let us write the appropriate equilibria and associate the correction <img src="https://tex.z-dn.net/?f=K_b" id="TexFormula1" tit

rosijanka [135]

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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