A solution is made 1.1⨉10 -3 M in Zn(NO 3 ) 2 and 0.150 M in NH 3 . After the solution reaches equilibrium, what concentration of Zn2 + (aq) remains?

Hello. Everyone in this video. We're trying to determine the concentration of R N. I two plus. KD on in a solution. So let's go ahead and remind ourselves that this compound here is important to us because the N. I and oh three compounds can go ahead and associate or be consisted of R N. I two plus can ion and two moles of R N 03, poly atomic and ion. So let's go ahead now and construct our ice table. So we have our I. C. And E. The equation is going to be and I two plus mixing in order six moles of N. H three will have our equilibrium arrows. And then our product of N I N age 36 and that's going to be a two plus charge. Alright now plug in the values you see from here. So that's going to be 1.3 times 10 to the -3. Same goes in here and this we do not know As for NH three that's going to be right over here, we'll just write that initial polarity 0.173. Then we have six moles six times this right over here, 1.3 times 10 to the -3. That's going to give us a total of 0.1652. Initially we have zero Then our 1.3 times 10 to -3. Then of course that's going to equal to 1.3 times 10 to the -3. Alright, so my K F value then how we can find this is through the concentration of N. I. And age 362 plus Divided by concentration of n. i. two plus multiplied by the concentration of N age three. Of course to the power of six because there's six moles. Now using this equation, we can go ahead and put in our numerical values, we'll have our 1.3 times 10 to the -3. Let's go ahead and rewrite that three there. Alright. And that's going to divide it by we don't know the concentration of this, that's what we're trying to solve for. And of course we have this which is just 1.652 to the power of six. And putting that into my calculator, I'll get 1.2 times 10 to the 9th power solve it for X. That's just equal to 1.3 times 10 to the -3. To fight it by Our 1.2 times 10 to the night, which we just have software times are 0.1652 to the power of six. We're sorry for this right here right here. All right now putting all of this into my calculator, I'll get the final value of 5.3297 times 10 to the negative eight. And of course to make this six figure appropriate. Go ahead round after three and two is just less than five. So we'll just have 5.3. So the concentration of R. N. I. Two plus carry on. Which we just saw for right here is going to equal to 5. times 10 to the negative eight power. And this is going to be my final answer for this problem. Thank you all so much for watching.

Ch.18 - Aqueous Ionic Equilibrium

Tro - Chemistry: A Molecular Approach 6th Edition

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Ch.18 - Aqueous Ionic Equilibrium

Problem 115

Chapter 18, Problem 115

A solution is made 1.1⨉10^-3 M in Zn(NO₃)₂ and 0.150 M in NH₃. After the solution reaches equilibrium, what concentration of Zn2⁺(aq) remains?

Verified step by step guidance

Identify the relevant chemical equilibrium: The complexation reaction between Zn(NO_3)_2 and NH_3 forms a complex ion, typically Zn(NH_3)_4^{2+}. The equilibrium can be represented as: Zn^{2+} + 4NH_3 \rightleftharpoons Zn(NH_3)_4^{2+}.

Write the expression for the equilibrium constant (K_f) for the formation of the complex ion: K_f = \frac{[Zn(NH_3)_4^{2+}]}{[Zn^{2+}][NH_3]^4}.

Determine the initial concentrations: Initially, [Zn^{2+}] = 1.1 \times 10^{-3} M and [NH_3] = 0.150 M. Assume [Zn(NH_3)_4^{2+}] = 0 M initially.

Set up an ICE (Initial, Change, Equilibrium) table to track the changes in concentrations as the system reaches equilibrium. Define the change in concentration of Zn^{2+} as -x, NH_3 as -4x, and Zn(NH_3)_4^{2+} as +x.

Substitute the equilibrium concentrations into the K_f expression and solve for x, which represents the change in concentration of Zn^{2+}. The equilibrium concentration of Zn^{2+} will be the initial concentration minus x.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium in Chemical Reactions

Equilibrium in chemical reactions refers to the state where the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. In this context, understanding how the concentrations of Zn2+ and NH3 interact at equilibrium is crucial for determining the final concentration of Zn2+ in the solution.

Complex Ion Formation

Complex ion formation occurs when metal ions, such as Zn2+, interact with ligands like NH3 to form a complex. This process can significantly affect the concentration of free metal ions in solution, as the formation of complexes reduces the amount of free Zn2+ available, which is essential for calculating the equilibrium concentration.

Le Chatelier's Principle

Le Chatelier's Principle states that if a system at equilibrium is disturbed, the system will adjust to counteract the disturbance and restore a new equilibrium. In this scenario, changes in concentration of NH3 or Zn(NO3)2 will shift the equilibrium position, impacting the concentration of Zn2+ ions in the solution.