Question

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9. Given that the Ksp for calcium fluoride [CaF2] is 3.2 x 10^-15, which of the following describes a solution that is 2.00 x 10^-5 M NaF and 2.00 x 10^5 M Ca(NO3)2? A) Q = 4 x 10^-10 and there will be a precipitate formed B) Q = 8 x 10^-10 and there will be a precipitate formed C) Q = 8 x 10^-15 and there will be no precipitate formed D) Q = 1.3 x 10^-13 and there will be a precipitate formed E) Q = 8 x 10^-15 and there will be a precipitate formed 10. Which of the following leads to the formation of a precipitate? A)Q = Ksp B)Q < Ksp C)Q > Ksp D)Q = 1 E)K = 1

Answer 1

(9) Given K_sp of CaF₂ is = 3.2x10 ^-15

[NaF ] = 2.00x10 ^-5 M

[Ca(NO₃)₂ ] = 2.00x10 ^-5 M

                             NaF $\rightarrow$ Na⁺ + F^-

                     Ca(NO₃)₂$\rightarrow$ Ca²⁺ + 2NO₃^-

[Ca²⁺ ]= 2.00x10 ^-5 M & [ F^- ]= 2.00x10 ^-5 M

So ionic product of CaF₂ is Q = [Ca²⁺ ][ F^- ]²

                                           = (2.00x10 ^-5 )(2.00x10 ^-5 )²

                                           = 4 x10 ^-10   

$\therefore$ Q = 4 x10 ^-10   

Given K_sp = 3.2x10 ^-15

So Q < K_sp , precipitate will form.

(10)

If Q < K_sp reaction is going forward and products are produced that is precipitate will form
Q > K_sp reaction is going backward and reactants are produced
Q = K_sp The rate of forward and backward is the same    

So to form a precipitate Q < K_sp