What is the equilibrium concentration of ZN^₂+(aq) in a solution that is initially 0.0100 M solution in Zn(NO₃) ₂ at a pH of 13.00?
Zn²⁺(aq) + 4 OH ^- (aq) fi Zn(OH) ₄² ^- (aq) K_f=

Question

What is the equilibrium concentration of ZN^₂+(aq) in a solution that is initially 0.0100 M solution in Zn(NO₃)₂ at a pH of 13.00? Zn²⁺(aq) + 4 OH ^- (aq) fi Zn(OH)₄² ^- (aq) K_f=

Quizplus · Accepted Answer

At pH 13, [OH⁻] is 0.1 M. Using the formation constant Kf = 2.5 x 10¹⁵, the equilibrium concentration of Zn(OH)₄²⁻ can be calculated. The high Kf indicates nearly complete formation of Zn(OH)₄²⁻, resulting in an equilibrium concentration close to the initial Zn²⁺ concentration, adjusted for stoichiometry and [OH⁻].

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