Respuesta :
Answer:
53.18 gL⁻¹
Explanation:
Given that:
[tex]Cu^{2+}_{(aq)}[/tex] [tex]+[/tex] [tex]2NH_{3(aq)}[/tex] [tex]------>[/tex] [tex][Cu(NH_3)_2]^+_{(aq)}[/tex] ------equation (1)
where;
Formation Constant [tex](k_f) =[/tex] [tex]6.3*10^{10}[/tex]
However, the Dissociation of [tex]CuBr_{(s)[/tex] yields:
[tex]CuBr_{(s)}[/tex] ⇄ [tex]Cu^{+}_{(aq)}[/tex] [tex]+[/tex] [tex]Br^-_{(aq)}[/tex] -------------- equation (2)
where;
the Solubility Constant [tex](k_{sp})[/tex] [tex]= 6.3 *10^{-9[/tex]
From equation (1);
[tex](k_f) =[/tex] [tex]\frac{[[Cu(NH_3)_2]^+]}{[Cu^{2+}][NH_3]^{2}}[/tex] --------- equation (3)
From equation (2)
[tex](k_{sp})[/tex] [tex]= [Cu^+][Br^-][/tex] --------- equation (4)
In [tex]NH_3[/tex], the net reaction for [tex]CuBr_{(s)[/tex] can be illustrated as:
[tex]CuBr_{(s)[/tex] [tex]+[/tex] [tex]2NH_{3(aq)}[/tex] ⇄ [tex][Cu(NH_3)_2]^+_{(aq)}[/tex] [tex]+ Br^-_{(aq)}[/tex]
The equilibrium constant (K) can be written as :
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}[/tex]
If we multiply both the numerator and the denominator with [tex][Cu^+][/tex] ; we have:
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}*\frac{[Cu^+]}{[Cu^+]}[/tex]
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+}{[NH_3]^2[Cu^+]}*{[Cu^+][Br^-]}[/tex]
[tex]K = k_f *k_{sp}[/tex]
[tex]K= (6.3*10^{10})*(6.3*10^{-9})[/tex]
[tex]K= 3.97*10^2[/tex]
[tex]K[/tex] ≅ [tex]4.0*10^2[/tex]
Now; we can re-write our equilibrium constant again as:
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}[/tex]
[tex]4.0*10^2 = \frac{(x)(x)}{(0.76-2x)^2}[/tex]
[tex]4.0*10^2 = \frac{(x)^2}{(0.76-2x)^2}[/tex]
[tex]4.0*10^2 = (\frac{(x)}{(0.76-2x)})^2[/tex]
By finding the square of both sides, we have
[tex]\sqrt {4.0*10^2} = \sqrt {(\frac{(x)}{(0.76-2x)})^2[/tex]
[tex]2.0*10 = \frac{x}{(0.76-2x)}[/tex]
[tex]20(0.76-2x) =x[/tex]
[tex]15.2 -40x=x[/tex]
[tex]15.2 = 40x +x[/tex]
[tex]15.2 = 41x[/tex]
[tex]x = \frac{15.2}{41}[/tex]
[tex]x = 0.3707 M[/tex]
In gL⁻¹; the solubility of [tex]CuBr_{(s)[/tex] in 0.76 M [tex]NH_3[/tex] solution will be:
[tex]= \frac{0.3707 mole of CuBr}{1L}*\frac{143.45 g}{mole of CuBr}[/tex]
= 53.18 gL⁻¹
The solubility of the CuBr in 0.76 M ammonia has been 58.18 g/L.
The equation for the solubility of CuBr can be given as:
[tex]\rm Cu^2^+\;+\;2\;NH_3\;\rightarrow\;[Cu(NH_3)_2]^+[/tex]
kf for the reaction has been:
kf = [tex]\rm \dfrac{[Cu(NH_3)_2]^+}{[Cu^+]\;[NH_3]^2}[/tex]
The copper ions have been formed from the dissociation of CuBr as:
[tex]\rm CuBr\;\leftrightharpoons \;Cu^+\;+\;Br^-[/tex]
The dissociation constant for the reaction ksp can be given as:
ksp = [tex]\rm [Cu]^+\;[Br]^-[/tex]
The net equation for the dissolution of CuBr with ammonia can be given as:
[tex]\rm CuBr\;+\;2\;NH_3\;\rightarrow\;[Cu(NH_3)_2]^+\;+\;Br^-[/tex]
The equilibrium constant for the reaction has been given as:
K = [tex]\rm\dfrac{ [Cu(NH_3)_2]^+\;[Br^-]}{[NH_3]}[/tex]
By multiplying both numerator and denominator by [[tex]\rm Cu^+[/tex]]:
K = [tex]\rm\dfrac{ [Cu(NH_3)_2]^}{[NH_3]\;[Cu^+]}\;[Cu^+]\;[Br^-][/tex]
K = kf × ksp
K = 6.3 [tex]\rm \times\;10^1^0[/tex] × 6.3 [tex]\rm \times\;10^-^9[/tex]
K = 397
K = 400
Let the equilibrium concentration of copper ammonium ion and bromide ion be x. The equilibrium concentration of ammonia has been 0.76 M.
K = [tex]\rm \dfrac{(x)\;(x)}{0.76}[/tex]
400 = [tex]\rm \dfrac{(x)\;(x)}{0.76}[/tex]
x = 0.3707 M.
The concentration of Br ion has been 0.3707 M. In CuBr the concentration of Br ion has been same as the concentration of CuBr.
The solubility of CuBr has been 0.3707 mol/L.
The solubility of CuBr has been 0.3707 × 143.45 g/L
The solubility of CuBr has been = 58.18 g/L.
The solubility of the CuBr in 0.76 M ammonia has been 58.18 g/L.
For more information about the solubility of the compound, refer to the link:
https://brainly.com/question/7176737
