A solution is prepared by adding 0.18 mole ofNi (NH3)6 Cl2 to 0.50 L of 2.0 MNH3. Calculate[Ni (NH3)6 ²+ ] and[Ni²+] in this solution.Koverall for2+Ni (NH3)6 is5.5 × 108. That is,5.5 x 108 -for the overall reactionNi²+ (aq) + 6NH3(aq) ⇒ Ni (NH3)²+ (aq)Ni (NH3)6²+][N₁²+] =2+[Ni (NH3),²+][Ni2+ ] [NH3]®16=Submit AnswerMMTry Another Version1 item attempt remaining

Answer:Transition metal ions have the tendency to form a very stable complex with electron rich…

A solution is prepared by adding 0.18 mole of Ni(NH3) 6 Cl₂ to 0.50 L of 2.0 M NH3. Calculate [Ni(NH3)²+] and Ni²+ in this solution. 2+ Koverall for 2+ Ni (NH3)²+ is 5.5 × 108. That is, 5.5 × 108 [Ni²+] = = Ni (NH3)2+ [Ni²+ ][NH® for the overall reaction 2+ Ni²+ (aq) + 6NH3 (aq) — Ni (NH3)6²+ (aq) 2+ [Ni (NH3)6 ²+] = M M

A solution is prepared by adding 0.18 mole of Ni(NH3) 6 Cl₂ to 0.50 L of 2.0 M NH3. Calculate [Ni(NH3)²+] and Ni²+ in this solution. 2+ Koverall for 2+ Ni (NH3)²+ is 5.5 × 108. That is, 5.5 × 108 [Ni²+] = = Ni (NH3)2+ [Ni²+ ][NH® for the overall reaction 2+ Ni²+ (aq) + 6NH3 (aq) — Ni (NH3)6²+ (aq) 2+ [Ni (NH3)6 ²+] = M M

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

A solution is prepared by adding 0.18 mole of
Ni (NH3)6 Cl2 to 0.50 L of 2.0 M
NH3. Calculate
[Ni (NH3)6 ²+ ] and
[Ni²+] in this solution.
Koverall for
2+
Ni (NH3)6 is
5.5 × 108. That is,
5.5 x 108 -
for the overall reaction
Ni²+ (aq) + 6NH3(aq) ⇒ Ni (NH3)²+ (aq)
Ni (NH3)6²+]
[N₁²+] =
2+
[Ni (NH3),²+]
[Ni2+ ] [NH3]®
16
=
Submit Answer
M
M
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Step 1: Formation of complex ion!

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Step 2: Calculation for concentration of complex ion initially!

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Step 3: Formation of ICE table

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