answer IV and V with correct units. please be detail (use the information below

Question

answer IV and V with correct units. please be detail (use the information below for problem 2)

  

What is/are the effect of addition of sodium bromide on the magnitude of the equilibrium constant of lead (IV) bromide?(you must fully discuss your answer before you can receive full credit) V. Assuming the reaction is exothermic, what is the effect of: A. Increase in temperature on magnitude of the reaction equilibrium constant of lead (IV) bromide? B. Decrease in temperature on magnitude of the reaction equilibrium constant of lead (IV) bromide? Lead (IV) bromide is a sparingly soluble salt in aqueous solution and dissociates to its respective ions in an equilibrium reaction. Write the solubility product (K_sp) expression for the lead (IV) bromide salt What is the concentration of lead ions if the solubility product of lead (IV) bromide salt is 3.5 times 10 s and the concentration of bromide ions is 3 mM? What is/are the effects of addition of calcium sulfate salt on the equilibrium constant of lead (IV) bromide? (you must fully discuss your answer before you can receive full credit)

Explanation / Answer

IV. Addition of sodium bromide would reduce the equilibrium constant of PbBr4 salt. NaBr completely dissociate into Na+ and Br- ions. This increases the Br- concentration in solution and thus disscoiation of PbBr4 is lowered. The increased Br- would combine with Pb4+ in solution to form more undissociated PbBr4 salt.

V. If the reaction is exothermic,

A. increase in temperature would reduce the equilibrium constant of PbBr4 as the reaction would try to bring the temperature down to the original equilibrium state by forming more undissociated PbBr4. That is reverse reaction would be favoured.

B. A decrease in temperature would increase the equlibrium constant as equilibrium will shift towards the product end.

2. For lead(IV) bromide

I. Ksp for lead(IV) bromide can be written as,

Ksp = [Pb4+][Br-]^4

II. with Ksp = 3.5 x 10^-5

Solubility of [Pb2+] = Ksp/[Br-]^4

with,

[Br-] = 0.003 M

we get,

[Pb2+] = 3.5 x 10^-5/(0.003)^4

            = 4.32 x 10^5 M

III. CaSO4 addition would decrease the equilibrium constant of PbBr4

CaSO4 when added, would saturate the solution with its ions and thus less amount of PbBr4 would dissociate into its ions, which in turn would reduce the equilibrium constant for PbBr4 in solution.

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