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?

Video transcript

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.