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What is the conjugate acid of each of the following? What is the conjugate base of each?$$\begin{array}{l}{\text { (a) } \mathrm{OH}^{-}} \\ {\text { (b) } \mathrm{H}_{2} \mathrm{O}} \\ {\text { (c) } \mathrm{HCO}_{3}^{-}} \\ {\text { (d) } \mathrm{NH}_{3}} \\ {\text { (e) } \mathrm{HSO}_{4}-} \\ {\text { (f) } \mathrm{H}_{2} \mathrm{O}_{2}} \\ {\text { (g) } \mathrm{HS}^{-}} \\ {\text { (h) } \mathrm{H}_{5} \mathrm{N}_{2}^{+}}\end{array}$$

a) Conjugated acid of $O H^{-}$ is $H_{2} O$Conjugated base of $O H^{-}$ is $O^{2-} .$b) Conjugated acid of $H_{2} O$ is $H_{3} O^{+}$Conjugated base of $H_{2} O$ is $O H^{-}$c) Conjugated acid is of $H C O_{3}^{-}$ is $H_{2} C O_{3}$Conjugated base is $C O_{3}^{2-}$d) Conjugated acid is $N H_{4}^{+}$Conjugated base is $N H_{2}^{-}$e) Conjugated acid is $H_{2} S O_{4}$Conjugated base is $S O_{4}^{2-}$f) Conjugated acid is $H_{3} O_{2}^{+}$Conjugated base is $H O_{2}^{-}$g) Conjugated acid is $H_{2} S$Conjugated base is $S^{2-}$h) Conjugated acid is $N_{2} H_{6}^{2+}$Conjugated base is $N_{2} H_{4}$

Chemistry 102

Chapter 14

Acid-Base Equilibria

Liquids

Carleton College

Drexel University

University of Maryland - University College

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problem. Seven from Chapter 14 is asking us to take the molecules that are provided in the question and, uh, give the conjugated asset and base forms of that molecule. So what that means in terms of a conjugated acid, it means that the originally starting molecule must have accepted a proton, and the resulting molecule now has an additional proton. Where and now it has become acidic and that it could lose that proton again in a reverse reaction, The opposite applies for a conjugal base. The original molecule to become the conjugal base must have lost a proton. So the cognac it base is, um, capable of accepting a proton. Um, which is why it is the conjugal base. So let's start. Bye. Um, let's start by answering the question A You're part of the question. So you have ah hydroxide ion. And on one side let's, um this is going to be, um, the addition of a hydrogen. I'll addition of a proton. This molecule on this side is going to be the conjuring it acid, which in this case is water. And when we convert this molecule into its conjugated based form, we are subtracting a proton and we are left with the oxygen atom. Great. So let's move on to part B of this question where we have bicarbonate and just to save, um, some time I'm not going to add the additional proton on top of this side. Just know that everything to the right is going to be the contra get acid form of the molecule. So when we add a proton to bicarbonate, we're going to get carbonic acid. So gains a proton, converting it to carbonic acid. And then when we lose a proton, we are going to get C 03 to Linus again, I'm not going to write the, uh, loss of a proton on the left hand side. So all the things in the left hand side are going to be the conjugal base From now on, next molecule we have is ammonia, and we have, um, we gain a proton. So then this ammonia is going to become ammonia. And on the other side you have a loss of proton, so it's going to become an age to minus, which is a contra get base form of ammonia. All right, The next molecule we have is hydrogen sulfate is gonna go on to the next page again. Right side's going to denote the, um, conjuring acid. So here we are, gaining a proton. So this is going to become sulfuric acid, H two s 04 And on the other side, we are losing a hydrogen ion, and we are left with so fate. Okay, next we have hydrogen peroxide and it's congregate. Acid form is going to be a TSH three. So that H two comes in h three with the additional proton, and this is going to have a positive charge. Well, rights, On the other hand, we are losing a proton to form a conjugal base of this H. The hydrogen peroxide is going to turn into water. Look. Oh, sorry. It's not gonna turn into water. It is going to be a choux to minus waters age to, um next we have Bye, so fine. H s minus. And we're going to add a proton to form the conjugal assets that gives us a TSH to s. And on the other side, we're going to turn it into a con. Trick it base. So we lose this only hydrogen ion that we have and that because the sulfide ion and lastly we have 86 Excuse me, 85 and two plus h five and two plus. So we ah, on the right hand side. So we are gaining a proton to become a contra Get acid. So this h is now going to be in each six and two. And now we have two plus because we gained a proton. Now on the left hand side, we are losing a proton. So instead of h five is going to be aged for each four into with no charge because we lost the positive charge when we lost a proton. So that is the answer for Question seven.

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