Using dissociation constants from Appendix D, determine the value...

Question

Using dissociation constants from Appendix D, determine the value for the equilibrium constant for each of the following reactions.
(i) HCO3-1aq2 + OH-1aq2 ΔCO32-1aq2 + H2O1l2
(ii) NH4+1aq2 + CO32-1aq2 ΔNH31aq2 + HCO3-1aq2

Accepted Answer

Hello everyone. So in this video we're trying to determine the value of the equilibrium constant for the following reactions. Using our dissociation constants. So we have part a and part to be here let's go ahead and do the first one. So we're gonna go ahead and first right the reactions for h starting material and then we'll go ahead and do the math. So starting off with our first agent we have H. C. R. 04 minus. That's going to go ahead and associate into C. R. 042 minus and R. H. Plus. Then for O. H. That's going to go ahead and react with R. H. Plus to go ahead and give us our H. +20. The fallen K values which also known as K one for the first reaction and K two for the second. That just comes from values that I found in my textbook is either in your textbook as well or given to you by your professor. So K one for this is gonna be 3.0 times 10 to the negative seven. And the K two for this is going to be one over my K. W. Which is going to equal to one times 10 to the positive 14 because we have one over the K. W. Alright then for our K value, The equation for that is going to be K1 times RK two And that just is 3.0 times 10 to the negative seven multiplied by one times 10 to the 14. Putting that into my calculator I'll get the final value of three times 10 to the positive seven power. Okay so that's part one to answer. Okay next we'll do part two. So again same exact format that we did up there. We'll start off with our first starting reagent C. Six, H. Five and H. Three. That's going to go ahead and associate into C. Six, H. Five and H. Two and R. H. Plus. Then our second star agent we have H. S. Minus reacting with H. Plus to go ahead and give us a church to s going ahead to write the K. Values down. We have K. One again and K. Two K one is equal to 2.56 times 10 to the negative five K. Two is going to be one over the K. Value Of our H. two s. And that is going to be 1.124 times 10 to the positive seven power. Just like what we did for the first part, we're gonna go ahead and multiply our K. One value and K. Two value to give us our K. Value which is our final answer. So plugging in those numerical values, We have 2.56 times 10 to the negative five, multiplying with 1.124 times 10 to the positive seven power. Putting that value into my calculator. My final answer is going to be 2.88 times to the positive second power. So this is going to be my answer for part two again, this is my K value for the first part here, and this is going to move the value for the second part. Thank you all so much for watching.

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