Question

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?

Answer 1

Answer:

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:

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

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

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

$[Br-] = 2.38 M$

Answer 2

The study of chemicals is called chemistry. when the amount of reactant or product gets equal is said to be an equilibrium state.

The correct answer is 2.38M.

What is equilibrium constant?

• The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium. A state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.

The data is given in the question is as follows:-

$Kf =5*10^3\\Ksp =2.5*10^{-14}$

The reaction in the given question is as follows:-

$Cd(OH)_2 +4Br^- +[CdbBr_4]^{2-} +2OH^-$

The formula we used to solve the question is as follows:-

$K =\frac{{[cdbr_4^2-]}[OH]^2}{{Br^-}^2}$

After placing the value, the correct answer for the bromine is 2.38M.

Hence, the correct answer is 2.38M

For more information about the equilibrium constant, refer to the link;-https://brainly.com/question/17960050