Respuesta :
Explanation:
The reaction equation will be as follows.
[tex]NH_{3}.H_{2}O(aq) + MgCl_{2}(aq) \rightleftharpoons Mg(OH)_{2} + NH_{4}Cl[/tex]
[tex]K_{sp}[/tex] of [tex]Mg(OH)_{2}[/tex] = 13 and [tex]K_{b}[/tex] of [tex]NH_{3}[/tex] = [tex]10^{-4.75}[/tex]
Hence, concentration of [tex]MgCl_{2}[/tex] = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
= [tex]\frac{0.2 mol}{50 \times 10^{-3}}[/tex]
= 4 M
As, [tex]Mg(OH)_{2} \rightleftharpoons Mg^{2+} + 2[OH^{-}][/tex]
Hence, [tex]K_{sp} = [Mg^{2+}][OH^{-}]^{2}[/tex]
13 = [tex]4 \times [OH^{-}]^{2}[/tex]
[tex][OH^{-}][/tex] = 1.80276 M
Also, ionization constant of water is given as follows.
[tex]K_{w} = [H_{3}O^{+}][OH^{-}][/tex] ......... (1)
[tex]K_{w} = 10^{-14}[/tex]
Now, put the value of [tex][OH^{-}][/tex] in equation (1) as follows.
[tex]K_{w} = [H_{3}O^{+}][OH^{-}][/tex]
[tex][OH^{-}] = \frac{K_{w}}{[H_{3}O^{+}]}[/tex]
= [tex]\frac{10^{-14}}{1.80276}[/tex]
= [tex]0.554 \times 10^{-14} M[/tex]
Also, [tex]K_{w} = K_{a} \times K_{b}[/tex] = [tex]10^{-14}[/tex]
Therefore, [tex]K_{a}[/tex] = [tex]\frac{10^{-14}}{k_{b}}[/tex]
= [tex]\frac{10^{-14}}{10^{-4.75}}[/tex]
= [tex]10^{-9.75}[/tex]
Since, [tex]H_{2}O(l) + NH^{+}_{4} \overset{K_{a}}{\rightleftharpoons} H_{3}O^{+}(aq) + NH_{3}(aq)[/tex]
[tex]K_{a} = \frac{[H_{3}O^{+}[NH_{3}]]}{[NH^{+}_{4}]}[/tex] ....... (2)
Hence, calculate the value of [tex][NH^{+}_{4}][/tex] as follows.
[tex][NH^{+}_{4}][/tex] = [tex]\frac{[H_{3}O^{+}][NH_{3}]}{K_{a}}[/tex]
= [tex]\frac{0.554 \times 10^{-14} \times 1}{10^{-9.75}}[/tex]
= [tex]0.554 \times 10^{-4.75} M[/tex]
[tex][NH^{+}_{4}][/tex] = [tex]3.115 \times 10^{-5} M[/tex]
Thus, we can conclude that the minimum concentration of [tex][NH^{+}_{4}][/tex] that prevents the precipitation of [tex]Mg(OH)_{2}[/tex] is [tex]3.115 \times 10^{-5} M[/tex].
