Predict the products of the following acid-base reactions,...

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the reaction arrow: (a) $\mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons$ (b) $\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)$ (c) $\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}(a q) \rightleftharpoons$

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the reaction arrow:
(a) $\mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons$
(b) $\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)$
(c) $\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}(a q) \rightleftharpoons$

Key Concepts

Bronsted-Lowry Acid-Base Theory

This concept defines acids as proton donors and bases as proton acceptors. It establishes the framework for acid-base reactions by focusing on the transfer of protons between reactants, leading to the formation of conjugate acid-base pairs. Understanding this theory is essential for predicting reaction products in acid-base chemistry.

Conjugate Acid-Base Pairs

In any acid-base reaction, the reactants and products can be viewed as conjugate pairs. The acid donates a proton to form its conjugate base, and the base accepts a proton to form its conjugate acid. Recognizing these pairs helps in determining the direction of the reaction and comparing the relative strengths of acids and bases.

Acid and Base Strengths

The strength of an acid or a base is determined by its tendency to donate or accept a proton, respectively. Strong acids and bases dissociate nearly completely in solution, while weak ones do not. Comparing the strengths of the acids and bases involved in a reaction is crucial in predicting the direction in which the equilibrium will lie, with the reaction generally favoring the formation of the weaker acid and base.

Chemical Equilibrium

Chemical equilibrium in the context of acid-base reactions refers to the point at which the rate of the forward reaction equals the rate of the reverse reaction. The position of equilibrium is influenced by the relative stability and strength of the acid-base pairs involved. By evaluating these factors, one can predict whether the equilibrium will lie to the left (favoring reactants) or to the right (favoring products).

Transcript

00:01 All right, we're going to be doing chapter 16, problem 26, from chemistry, the central of science, by brown, the 14th edition.

00:10 And the question is asking us to predict the products of the following acid -based reactions and predict whether equally bromeralized to the left or to the right of the reaction.

00:19 So the first thing we're going to have to do for each other questions is finish off the reaction.

00:26 So first we have an h4 plus plus oh minus and that is going to go to nh3 because that oh is going to take the hydrogen from the nh4 plus so therefore we have an h3 plus an h2o.

00:51 And then for the next problem we're going to just finish the reaction to for all of them.

00:58 We have a ch3co minus plus h3o plus.

01:02 And what's going to happen is that a hydrogen from this h3o is going to be added to the hco minus.

01:11 So now we're going to have ch3coooh plus h2o.

01:26 And then now for the last one, we're going to have an hco3 -equius plus f2o.

01:27 And then now for the last one, we're going to have an hco3.

01:35 And then what's going to happen is that this h is going to be added to that f right here.

01:41 And so now we're going to have the co3 -2 -1 plus h.

01:55 All right, since now we have all the reactions finished.

01:58 Now we can figure out which ones, what are we going to use to compare in the equation to be able to figure out which one's going, which way it's going to shift.

02:06 So the best way to do this is to figure out which one is the strongest.

02:10 And then from there it is going to be pushing the opposite way.

02:15 And so in this equation, we automatically can see that there's an oh in this equation.

02:20 And then we can figure out what is its conjugate acid or conjugate base equation.

02:25 So we have an h4 plus and oh minus.

02:28 We know that this is a base and that this is the acid.

02:32 And therefore, we have a conjugate baser here.

02:36 And it's a conjugate acid over here.

02:40 And so whenever you're looking at the oh minus, this is a stronger base than the conjugate base on the other side of nh3.

02:48 And so since this oh minus is stronger, the reaction is going to rely to the right...

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