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23.What Is the etfect on the equlllbrlum when sodlum formate Is added t0 solutlon of formic acld? HCOOH(ad) = == HT(ad) HCOO-(aq)There is no change In the equilibri...

Question

23.What Is the etfect on the equlllbrlum when sodlum formate Is added t0 solutlon of formic acld? HCOOH(ad) = == HT(ad) HCOO-(aq)There is no change In the equilibriumThe equllibrium shifts to the rightMore information is needed t0 answer the question_The equilibrium shifts t0 the leftThe pH decreases

23. What Is the etfect on the equlllbrlum when sodlum formate Is added t0 solutlon of formic acld? HCOOH(ad) = == HT(ad) HCOO-(aq) There is no change In the equilibrium The equllibrium shifts to the right More information is needed t0 answer the question_ The equilibrium shifts t0 the left The pH decreases



Answers

Consider the equilibrium
$$\mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q)$$
Suppose that a salt of $\mathrm{HB}^{+}(a q)$ is added to a solution of $\mathrm{B}(a q)$ at equilibrium. (a) Will the equilibrium constant for the reaction increase, decrease, or stay the same? (b) Will the concentration of $\mathrm{B}(a q)$ increase, decrease, or stay the same? (c) Will the $\mathrm{pH}$ of the solution increase, decrease, or stay the same?

This question has been given to reactions H. Two S and H plus from that it's mine. We also have another reaction which we have this equilibrium and as to money. So I've been given the Casey for Of these reactions being equal to 9.5 bite into the bar -8. And we have Casey which is also for the second step which is also one bike 10 to the power negative one night. Now we've been given another reaction in which would have H. Two S. That is at equilibrium with two H. Plus and S two months. And we've been asked to determine the equilibrium constant. This Pat reaction. Now if we're not is we can tell that this that reaction is overall reaction of two steps here. So we have this H. S. That is going to cancel out with this one. So at the end of the day we're going to have a chess reacted and two moles of H plus protect plus another more of As to minors. And this is exactly what we have here in 30 years. So if we want to determine the K. C. Of this reaction. Now that reaction this is going to be. Since this is just some of these steps. What we can do is to just multiply the Casey values of the individual steps. So all K. C. Is going to be equal to Casey of the first reaction and multiplied by the Casey of the second reaction. So this gives us 9.5 multiplied by 10 to departing All of these, multiplied by one x 10 to our negative one night. So the K. C. Of the state reaction that we've been given This is equal to 9.5. I tend to depart -1.

And this problem, we are given the equilibrium constants, Casey for each one of these two reactions. And we want to determine what the equilibrium constant Casey is for the target reaction given in the problem statement, we examined the two reactions that were given. We should be able to see that in order to get to that target reaction, we have to add these two reactions together. So when we do that, we see that HS minus equals cancels out on each side, and the two H plus a quickest ions combine to give us two each plus as a product. And now we can see that when we write out the overall reaction each to S. Equis going to to H plus equally ists plus as to minus Equus. But that gives us the overall reaction that we want to calculate the value of K C for and whenever we add two reactions together like that. And we know the values of the equilibrium constants for each one of those reactions that in order to calculate the value of the equilibrium constant of the overall reaction, the equivalent to adding those chemical reactions in terms of equilibrium, constants is multiplying them together. And so to determine that value of K C. For this overall target reaction that would just be equal. Now just be equal to K C. Prime Times K C double Prime Casey Prime is 9.5 times 10 to the negative, eighth in K C double prime is 1.0 times 10 to the negative 19th. So we just multiply those two equilibrium constants together to get the overall equilibrium constant, which comes out to 9.5 times 10 to the power of negative 27.

In this question, we want to find the ph of this solution. Were given K Eq H P R O does not ionizing enough to affect the ph So the key to this question is that cake, you only depends on species in the acreage solution. So normally when we write the generic equation for KQ equilibrium constant, we have the products on top and then we have the reactant on the bottom. But these need to be in the acquis phase. So in this case you can see that all of the reactions are not in the acquis phase. So we only have to take into account the product. So we can write that K. Q. Is equal to the products of the products of the reaction. So we have to multiply the concentration of the products together and we don't raise them to any story geometric coefficients because you can see that in the reaction, they don't have any coefficients in front of them. So we can write this and then we just need to take the cube root of cake, you to find the concentration of H plus, which is what we're interested in to find ph so we do the cube group of the cake, you and then we find this is the concentration and then we can solve for the ph the ph is equal to negative log of concentration of hydro nia my on or H plus. And then we end up getting the ph from that. So the key here is knowing that cake. You only depends on concentration of products and reactions in the crease phase and not a solid or liquid. And also we need to know how to calculate the ph and that it depends on the H plus concentrate.

We are continuing on with our work of the reactivity and the properties of the non metals. And so the first thing we're looking to do here is draw a table for the concentration of the species in the solution. So what we have is B R. Two and H two. That is an equilibrium with H plus at B R miners at h b r O. So we have the initial the changes equilibrium. So we've got the ice table. So initially we have c minus X. See bracket one minus X with water. We aren't filling that one in each plus. We've got zero addicts, C X, zero addicts C X. And again zero addicts C X. So from the table we can see that at equilibrium H plus is equal to see X. So therefore we can write RK equation or equilibrium constant, K is equal to h plugs, multiplied by br minus, multiplied by h b r O. So they are the product divided by br to the starting materials. So we sub in the algebraic formula from before. So we have C. X. Multiplied by C. X. Multiplied by C X. All divided by C. Multiplied by one minus X. And so we solve X. Where that is C. X. For H plus where the H plus concentration, that is equal to 1.6 times 10 to the minus three. Now what we can do with this information is plug it into our equation to determine the P. H. So we have the P. H. Is equal to the negative log of H. Plus. We plug in that valley that we just looked up and what we get is 2.97


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