Nicotine 1C10H14N22 can accept two protons because it has
two bas...

Question

Nicotine 1C10H14N22 can accept two protons because it has
two basic N atoms 1Kb1 = 1.0 * 10-6; Kb2 = 1.3 * 10-112.
Calculate the values of Ka for the conjugate acids
C10H14N2H+ and C10H14N2H22 + .

Accepted Answer

Hello. Everyone in this video we're talking about arginine which has a molecular formula of C. Six H. 14 and four and 02. It can receive two protons due to the presence of two basic items which is nitrogen were given the KB one and KB two value are being asked where the conjugate assets this one and this one. What are the values of R K. S. So let's go ahead and recall that kw It goes to K. A. Times KB. Since we're solving for K. Here in this problem, we can just isolate K by dividing both sides of the equation by KB. Making our new equation then equaling two K. A equals to K. W. Over K B. KB is given. And K W is already something that we know. Alright, so for the first conjugate acid will do C six H 15 N 402 plus. Alright so we know that this contract acid so is the consequence acid four C 6 h. 14 And 402. And again we're given two KB values. We have K. B. One and K. B. Two. We'll go ahead and use the KB one again. We're using the equation for K. A. Going to K. W over K. B. Alright, so I'm just gonna go ahead and plug in the values here. So again Kw something that already know that's one times 10 to the negative 14 and KB is given to us in the problem and that's 1. times 10 to the -5. Right? This little bit more clearly. Alright so plugging this into my calculator. I see that my K value equals 21 times 10 to the negative nine. So that's my K. Value from our first conjugate acid, C six H. 15 and 402 plus. And moving on for a second conjugate acid. So now we're dealing with C six H. 16 And 402 with a two plus charge. So this time of course we'll use our KB two value. But just for consistency sake we can go ahead and also recognize that this is the conjugate acid, four C six H. 15 and 402 plus again. The equation here that we're using is K A equals two Kw over KB. So now I'm just gonna go ahead and plug in my numerical values again. KW is something that we know that's one times 10 to the negative 14. Over. And then KB we're using KB two. So let's say that as well. So earlier we use KB one. Now we're using KB two. So from a problem we see that that value is 6.3 times 10 to the negative 13. So once I put this value into my calculator, I get out 0.01587. So now rounding this to the correct six figs as well as putting into scientific notation, we see that we get K. Being 1.6 times 10 To the -2. So this here is my last answer for this problem

Nicotine 1C10H14N22 can accept two protons because it has two basic N atoms 1Kb1 = 1.0 * 10-6; Kb2 = 1.3 * 10-112. Calculate the values of Ka for the conjugate acids C10H14N2H+ and C10H14N2H22 + .

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