In the presence of excess OH, the Zn2+ (aq) ion forms a hydroxide complex ion, Zn(OH)42-. Calculate the concentration of free Zn2+ ion when 1.72x10-² mol Zn(CH3COO)2(s) is added to 1.00 L of solution in which [OH-] is held constant (buffered at pH 12.30). For Zn(OH)42-, Kf = 4.6x10¹7. [Zn²+] = M

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Chapter15: Acid-base Equilibria
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In the presence of excess OH, the Zn²+ (aq) ion forms a hydroxide complex ion,
Zn(OH)42. Calculate the concentration of free Zn2+ ion when 1.72x10-² mol
Zn(CH3COO)2(s) is added to 1.00 L of solution in which [OH-] is held constant
(buffered at pH 12.30). For Zn(OH)42-, Kf = 4.6x10¹7.
[Zn2+] =
M
Transcribed Image Text:In the presence of excess OH, the Zn²+ (aq) ion forms a hydroxide complex ion, Zn(OH)42. Calculate the concentration of free Zn2+ ion when 1.72x10-² mol Zn(CH3COO)2(s) is added to 1.00 L of solution in which [OH-] is held constant (buffered at pH 12.30). For Zn(OH)42-, Kf = 4.6x10¹7. [Zn2+] = M
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