Will a precipitate form when 100.0 mL of 6.8 x 10-4 M Mg(NO3)2 is added to...

Question

Question

Will a precipitate form when 100.0 mL of 6.8 x 10-4 M Mg(NO3)2 is added to 100.0 mL of 1.2 x 10-4 M NaOH? The ion product for

Answer 1

Answer #1

concentration of Mg2+ = 100 x 6.8 x 10^-4 / 100 + 100

= 3.4 x 10^-4 M

concentration of OH- = 100.0 x 1.2 x 10^-4 / 100 + 100

= 6.0 x 10^-5 M

Mg(OH)2 --------------> Mg2+ + 2 OH-

Q = [Mg2+][OH-]^2

= (3.4 x 10^-4) (6.0x 10^-5)^2

Q = 1.2 x 10^-12

the ion product for Mg(OH)2 is 1.2 x 10^-12 . since Q is less than Ksp.

Mg(OH)2 will not form precipitate.

Answer 2

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