Chemistry: The solubility of cd(oh)2 can be increased through formation of the complex ion cdbr2−4 (kf=5×103). if solid cd(oh)2 is added to a nabr solution, what would the initial concentration of nabr need to be in order to increase the molar solubility of cd(oh)2 to 1.0×10−3 moles per liter?

The solubility of cd(oh)2 can be increased through

Ответ:

kayciemarie9486

15.03.2022

Chemistry

Concentration of sodium bromide required = 2.38 M (around 2.4 M)

Explanation:

The equilibrium representing the complex ion formation is:

$Cd^{2+} + 4Br^{-}\rightleftharpoons [CdBr_{4}]^{2-} .....Kf =5*10^{3}$ -----(1)

where K(f) = formation equilibrium

The equilibrium representing the dissolution of Ca(OH)2 is:

$Cd(OH)_{2}\rightleftharpoons Cd^{2+}+2OH^{-}.....Ksp = 2.5*10^{-14}$ ---(2)

where Ksp = solubility product

adding Equation (1) and equation(2) gives the net reaction:

$Cd(OH)_{2} + 4Br^{-}\rightleftharpoons [CdBr_{4}]^{2-}+2OH^{-}$

$K = K_{f}*K_{sp} = 5*10^{3}*2.5*10^{-14}=\frac{[CdBr_{4}^{2-}][OH^{-}]^{2}}{[Br{-}]^{4}}$

$12.5*10^{-11} =\frac{1*10^{-3} *[2*10^{-3}]^{2}}{[Br-]^{4} }\\$

[Br-] = 2.38 M

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Ответ:

glowbaby123

27.04.2020

Chemistry

5) What is the net charge of every ionic compound? In every ionic compound, the total positive charge of the cations must always equal the total negative charge of the anions, so that the net charge of the complex is always zero

about question number 4 not having any idea

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The solubility of cd(oh)2 can be increased through formation of the complex ion cdbr2−4 (kf=5×103). if solid cd(oh)2 is added to a nabr solution, what would the initial concentration of nabr need to be in order to increase the molar solubility of cd(oh)2 to 1.0×10−3 moles per liter?

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