Ni forms a complex with ammonia (NH) which has the formula [Ni(NH) J, Calculate the [Ni). [NH3], and [Ni(NH) at equilibrium, when 0.300 M of [Ni*] and 1.200 M of [NH)] are present in the solution initially. [Kr for [Ni(NH)),- is

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Chapter15: Solubility And Complex Lon Equilibria
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Ni2* forms a complex with ammonia (NH3) which has the formula [Ni(NH:)6]?*.
Calculate the [Ni2*], [NH3], and [Ni(NH:)6"]´at equilibrium, when 0.300 M of [Ni²*]
and 1.200 M of [NH3] are present in the solution initially. [Kr for [Ni(NH3)6]?* is
5.5x10$]
Ni forms a complex with ammonia (NH) which has the formula [Ni(NH) J+ Calculate the [Ni). [NH), and [Ni(NH.)"]
at equilibrium, when 0.300 M of [Ni ] and 1.200 M of [NH] are present in the solution initially. [Kr for [Ni(NH.)F is
5.5x10 ]
Transcribed Image Text:Ni2* forms a complex with ammonia (NH3) which has the formula [Ni(NH:)6]?*. Calculate the [Ni2*], [NH3], and [Ni(NH:)6"]´at equilibrium, when 0.300 M of [Ni²*] and 1.200 M of [NH3] are present in the solution initially. [Kr for [Ni(NH3)6]?* is 5.5x10$] Ni forms a complex with ammonia (NH) which has the formula [Ni(NH) J+ Calculate the [Ni). [NH), and [Ni(NH.)"] at equilibrium, when 0.300 M of [Ni ] and 1.200 M of [NH] are present in the solution initially. [Kr for [Ni(NH.)F is 5.5x10 ]
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