Will a precipitate form if the pH of a 0.015 M solution of Ni(NO3)2 is adjusted to 9.00? Ksp Ni(OH)2 = 1.7 x10-6 .

Question

Question

Answer 1

Answer #1

pH = 9.00

pOH = 14 - pH

pOH = 14 - 9.00

pOH = 5.00

[OH^-] = 10^-pOH

[OH^-] = 10^-5.00

[OH^-] = 1.0 x 10^-5 M

initial concentration Ni²⁺ = [Ni(NO₃)₂] = 0.015 M

Reaction quotient, Q = [Ni²⁺][OH^-]²

Q = (0.015 M) * (1.0 x 10^-5 M)²

Q = 1.5 x 10^-12

Since Q < K_sp, therefore, precipitate will not form.

Answer 2

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