Respuesta :
Explanation:
Reaction equation is as follows.
[tex]Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)[/tex]
Here, 1 mole of [tex]Na_{2}SO_{3}[/tex] produces 2 moles of cations.
[tex][Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58[/tex]
= 1.16 M
[tex][SO^{2-}_{3}] = [Na_{2}SO_{3}][/tex] = 0.58 M
The sulphite anion will act as a base and react with [tex]H_{2}O[/tex] to form [tex]HSO^{-}_{3}[/tex] and [tex]OH^{-}[/tex].
As, [tex]K_{b} = \frac{K_{w}}{K_{a_{2}}}[/tex]
= [tex]\frac{10^{-14}}{6.3 \times 10^{-8}}[/tex]
= [tex]1.59 \times 10^{-7}[/tex]
According to the ICE table for the given reaction,
[tex]SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}[/tex]
Initial: 0.58 0 0
Change: -x +x +x
Equilibrium: 0.58 - x x x
So,
[tex]K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}[/tex]
[tex]1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}[/tex]
[tex]x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)[/tex]
x = 0.0003 M
So, x = [tex][HSO^{-}_{3}] = [OH^{-}][/tex] = 0.0003 M
[tex][SO^{2-}_{3}][/tex] = 0.58 - 0.0003
= 0.579 M
Now, we will use [tex][HSO^{-}_{3}][/tex] = 0.0003 M
The reaction will be as follows.
[tex]HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}[/tex]
Initial: 0.0003
Equilibrium: 0.0003 - x x x
[tex]K_{b} = \frac{x^{2}}{0.0003 - x}[/tex]
[tex]K_{b} = \frac{K_{w}}{K_{a_{1}}}[/tex]
= [tex]\frac{10^{-14}}{1.4 \times 10^{-2}}[/tex]
= [tex]7.14 \times 10^{-13}[/tex]
Therefore, [tex]7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}[/tex]
As, x <<<< 0.0003. So, we can neglect x.
Therefore, [tex]x^{2} = 7.14 \times 10^{-13} \times 0.0003[/tex]
= [tex]0.00214 \times 10^{-13}[/tex]
x = [tex]0.0146 \times 10^{-6}[/tex]
x = [tex][OH^{-}] = [H_{2}SO_{3}][/tex] = [tex]1.46 \times 10^{-8}[/tex]
[tex][H^{+}] = \frac{10^{-14}}{[OH^{-}]}[/tex]
= [tex]\frac{10^{-14}}{0.0003}[/tex]
= [tex]3.33 \times 10^{-11}[/tex] M
Thus, we can conclude that the concentration of spectator ion is [tex]3.33 \times 10^{-11}[/tex] M.
Otras preguntas
