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What is the ionization constant at $25^{\circ} \mathrm{C}$ for the weak acid $\mathrm{CH}_{3} \mathrm{NH}_{3}^{+},$ the conjugate acid of the weak base $\mathrm{CH}_{3} \mathrm{NH}_{2}, K_{\mathrm{b}}=4.4 \times 10^{-4}$

$K_{a}=2.3 \cdot 10^{-11}$

Chemistry 102

Chapter 14

Acid-Base Equilibria

Liquids

University of Central Florida

Rice University

University of Toronto

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problem. 35 from Chapter 14 is asking us to calculate the K a massive ionization constants the C h, the image you plus, which is the consequent acid to the weak base metal. I mean, we're giving the K B or based ionization constants of metal on me, which is 4.4 times 10 to the negative four. Um, so we know that k a k b equal k w We'll write that out. So be acid ionization constant multiplied by the base ionization constant is equal to kw or one times 10 to the minus 14. So all we really need to do to answer this question to solve for K A is just rearranges equation, um, in that will give us, um, kw divided by que me is going to equal. Okay, So let's go ahead and put the values and that we were given. So one times 10 to the minus 14 divided by 4.4 times 10 to the minus four. And when we do that, we are given a K A of 2.3 times 10 to the minus 11 and that is our kay

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