Use the appropriate values of Ksp and Kf to find the equilibrium constant for the reaction. FeS(s) + 6 CN-(aq)ΔFe(CN)64 - (aq) + S2 - (aq)

Video transcript

Hello everyone in this video we're dealing with k S p and k F values. So let's go ahead and write the equilibrium equations first. So we have this solid right here. So C U S of course. That's in solid form equilibrium arrows that can be see you two plus, which is actually acquis & S - which again is Aquarius and r k s. P value for this is good to be right over here. It's given to us. So RKSP value is going to be two times 10 to the -47 power. Then for our K. F value we'll have to see you two plus. That's going to be a chris it's going to react with our six moles of our cyanide and ion right over here and that's going to be Go ahead and add in those equilibrium arrows. And we have a product of CUCN 42 - and that is going to be a quiz. And that's how we have the K.F. given to us in the problem of one times 10 To be positive, 25 power. All right, so drawing, I like to separate our equations so I'm just gonna go ahead and rewrite out the given equation right over here. So we have our C U. S solid plus R six C n minus. That's going to be Aquarius equilibrium arrows will have c u c n 42 minus being our first product. And second product is our s two minus an eye on being acquis. Alright, so how we get the equilibrium constant value that's going to be K equals to r k S p. Multiplied by r k F value, scrolling down a little bit here, just following this equation then and give our plugging in our values that we are given to us right over here or in the problem. R k S p value as we can recall is two times 10 to negative 47. It's going to multiply bar one times 10 to 25th power. If I go ahead and put this into my calculator, I'll get the value of two times 10 to the negative 22nd power. And this is going to be my equilibrium constant value for the given reaction. Thank you all so much for watching.