Calculate Ka for the cation and Kb for the anion in an aqueous
N...

Question

Calculate Ka for the cation and Kb for the anion in an aqueous 
NH4CN solution. Is the solution acidic, basic, or neutral?

Accepted Answer

Hello everyone. So this video we're dealing with pure radium fluoride, which has a formula of C five H. Five N. H. F. So we're trying to determine the K. And the Kb of the canyon and an ion. So we want to go ahead and also identify of the resulting solution is acidic, basic or neutral. So we're given the K. Of H. F. And were given the KB of C five H five N. So something we just know from our common chemical knowledge is that C five H five and H F. Here this consist of C five H five N. H plus and F minus. So for our cat eye on here, this comes from a weak base this week base is C five H. 5 and H. And because of that this is actually acidic. And then we have our f minus here. This comes from a week acid. The acid is H. F. Which means here we have something that's basic. So for our cat on here, which is R C five H five N H. Plus we know that the equation for kW is equal to K. A. Times R K. B. Value here are solving for R. K. A. Because he said that this anti artist cat ion is acidic. We're gonna go ahead and divide both sides by KB two XLR K. If we do so we get the new equation that K A. Is equal to K. W over K. B. Of course we have these values already so we can go ahead, plug it in. So kwun something that we know that's equal to 1.0 times 10 to the negative 14 KB values. Something that were given to us in the problem and that is 1.8 times 10 to the negative nine. So once I put that into my calculator I see that my K. A value is equal to 5.6 times 10 to the negative six. As for are an eye on here which is f minus again. My kw equation is equal to K A. Times KB. Here, we're solving now for our KB value. So I'll divide both sides of the equation here by K. A. So I get that my new equation then to solve for K B is equal to K W over K A. So let's go ahead and plug in the values that we have here again two kw something that we know that's equal to 1.0 times 10 to the negative 14. For my K value, that's 3.5 times 10 to the negative four. So once I put that into my calculator I see that my KB value is equal to 2.9 times 10 to the negative 11. So here we have suffered R K A value of our caravan and KB value of my an ion. So now we can go ahead and look at these two values more closely. We see that my K value here is actually bigger than my KB. What this means is that the resulting solution is going to be acidic, since our K value is bigger. So we can say then that there are four, so that's three dots. The resulting solution is acidic, so this right here is going to be my final answer for this problem.

Calculate Ka for the cation and Kb for the anion in an aqueous NH4CN solution. Is the solution acidic, basic, or neutral?

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