Write a balanced net ionic equation for each of the follow-ing dissolution reactions, and use the appropriate Ksp and Kf values in Appendix C to calculate the equilibrium constant for each. (b) Cu(OH)2 in aqueous NH3 to form [Cu(NH3)4]2+

Video transcript

Hello everyone today. We have the following question provide the balanced non ionic equation for the dissolution reaction of nickel hydroxide and an Aquarius ammonia forming this nickel. This metal amine complex determine the equilibrium constant for the reaction, given that R. K. S. P. For our nickel hydroxide is 5.5 times 10 to the negative 16. And the K. F constant for this metal complex is two times 10 to the eighth. So first you want to draw our dissolution reaction of our nickel hydroxide. It's going to associate into one nickel to plus a crease ion as well as to hydroxide Aquarius ions. And we have that K. S. P. which was 5.5 times 10 to -16. So what we're gonna do is we're gonna take this nickel here, we're going to take that this nickel and we're going to react it with our I'm on you in the Equus form of course. And this is going to give us our metal, I mean, complex acquis ions of course. And to balance this, we see we have six ammonia on the right but one of the left, so we simply just have to add six in front of the ammonia on the left and this K. F. Value was two times 10 to the eighth. What we're gonna do is we're gonna try and cancel out any islands that appear twice on opposite sides. So essentially we can cross out our Nickel two plus because it's on the right of one equation and the left of the other. And now we can draw our overall reaction. So we have our nickel hydroxide is solid reacting with our six moles of our ammonia. This is going to give us our metal amine complex As well as our two hydroxide ions. And R K equilibrium is actually going to be the K. S. P times our K. F value. So with that equation we have our K. S. P times our K. F value. And so what's it gonna look like? What we're gonna have? Our 5.5 times 10 to the negative, 16th times our two times 10 to the eighth. R. K equilibrium or the equilibrium constant will be 1.1 times 10 to the negative seventh. And so we have our balanced equation and we also have our equilibrium constant here and with that we've answered the question overall, I hope this helped. And until next time.