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HAME AND STUDENT NUMBER-DO NOT Submit This PAGE[Alt = ~kt + [Alo In[AJt ~kt + In[Alokt + t1/2 = L [AJo ~6-4)t1/2 I2t1/2 = kiAlok = Ae%2 [Elol] Ky+[S]~b fvb_tac 1=ra...

Question

HAME AND STUDENT NUMBER-DO NOT Submit This PAGE[Alt = ~kt + [Alo In[AJt ~kt + In[Alokt + t1/2 = L [AJo ~6-4)t1/2 I2t1/2 = kiAlok = Ae%2 [Elol] Ky+[S]~b fvb_tac 1=ratepKa pKb = 14Kp = Kc(RT)AnPV = nRTpH = pKa[oase [acld] >5 =klnWASsuTT4,G" _ B,H" T4-S"ASsys ASsurt 4,G = 4,G" RT In Q4_HeIn K =46-0 4,G" ~RT In K4,G" = Evp4;G productsEvpA; reactanisA,s" = Evps? TroduclEvpS", reactantsA,H' = Evpb;H?F ooducEvp4;HE TeoctantsEocell Eocathodeniflndla~nFE&qu

HAME AND STUDENT NUMBER- DO NOT Submit This PAGE [Alt = ~kt + [Alo In[AJt ~kt + In[Alo kt + t1/2 = L [AJo ~6-4) t1/2 I2 t1/2 = kiAlo k = Ae %2 [Elol] Ky+[S] ~b fvb_tac 1= rate pKa pKb = 14 Kp = Kc(RT)An PV = nRT pH = pKa [oase [acld] > 5 =klnW ASsuTT 4,G" _ B,H" T4-S" ASsys ASsurt 4,G = 4,G" RT In Q 4_He In K = 46-0 4,G" ~RT In K 4,G" = Evp4;G products EvpA; reactanis A,s" = Evps? Troducl EvpS", reactants A,H' = Evpb;H?F ooduc Evp4;HE Teoctants Eocell Eocathode niflndla ~nFE" cell Ecell ln Q R = 8.314 ] mol-1K-1 0.0831 bar mol "K-1 k=138 * 10-23| K-1 T (K) = T(C) + 273 Kw = 1.0 * 10-14 96485 € mol-1 Page 13 of 14 iSunl" 4-G"



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The equilibrium equations and $K_{a}$ values for three reaction systems are given below. $\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}$ $$ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HC}_{2} \mathrm{O}_{4}^{-}(a q) ; K_{a}=5.6 \times 10^{-2} $$ $\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}=$ $$ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) ; K_{a}=6.9 \times 10^{-3} $$ $\mathrm{HCOOH}(a q)+\mathrm{H}_{2} \mathrm{O}=$ $$ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HCOO}^{-}(a q) ; K_{a}=1.7 \times 10^{-4} $$ a. Which conjugate pair would be best for preparing a buffer with a pH of $2.88$ ? b. How would you prepare $50 \mathrm{~mL}$ of a buffer with a pH of $2.88$ assuming that you had available $0.10 \mathrm{M}$ solutions of each pair?

This next question. Very similar to the previous question. I want you to determine which would serve best to achieve a ph Of 6.96. To answer this question, we need to calculate the P K. A values for the acids that are available. The one that has the peak value closest to the desired ph will be the one that will work best. The peak values can be calculated by taking the negative log of the K values and we get H to peel for minus Serbian is the best best acid to make a buffer at 6.96 having a PK value of 7.21. So to prepare this buffer solution with the target ph of 6.96 and a PK value of 7.21. The Henderson has the bolt equation states that ph equals Pekka plus the log of the moles of the base over the molds of the acid. The moles of the base will be equal to the volume of the base that we need to add. Which we don't know yet. Multiplied by its concentration of .1 Mueller. The most of the acid will be the volume of the acid that we need to add which we don't know yet But we want a total volume of 100 ml. So it'll be 100 mil leaders or 1000.1 liters minus the volume required for the base multiplied by the concentration of the acid. Now with this single algebraic expression We can simply rearrange it to solve for eggs will subtract 7.21 from both sides, take the anti log based tan on both sides, then multiply both sides by this denominator and collect our like terms, And then divide both sides by .1565, giving us 36 mil leaders of h p 04 to minus the base and 64 million liters of h two p 04 minus the acid.

So here's the question. 84 in Chapter 19 part a right, the chemical equation for the Taliban that was correspond to KP. First, we need to understand the term of K B, which means the association constant for bass. And in this case, is that the association off Ch three image too. So the reaction will be ch three in age, too. Chris Plus Water Liquid and the Associates into ch three Inish three plus hey Chris, Plus oh H minus also equals. And then part B using the value ofthe K B Cat Coy Delta zero. The relationship between Delta zero and KP is that Delta zero in for two negative Artie Long baby. And this is equal to negative. A point 314 times 10 to the minus. Three times 2 98 Kelvin Time's too long. 4.4 times 10 to the minus four and this is equal to 19.1 killed Jules. So that will be part B part. See, what is the value ofthe Delta G at equilibrium? So remember that as a hologram, the value ofthe Delta ji will always be zero now, the last one. What is the value ofthe Delta G when given the following conditions. So the relationship between Delta G and Delta zero is that Delta G is equal to zero plus Artie Long. Q. So the question tells you that the concentration of age plus is 6.7 times 10 to the minus nine. Moller, however, in the question, we were required to know the concentration of Polish miners, so the concentration of which minus and H plus there's a relationship between them that is the concentration of each plus times. The concentration of Polish minus will be equal to 10 to the minus 14. So in this case, for have the concentration off wish minus equal to 10 to the minus 14 divided by 6.7 times 10 to the minus night, and this will give you a number of 1.5 times 10 to the minus six. So now plugging the numbers to Q would have Delta G equal to Delta zero, which is 19.1 plus Artie. That is a 0.314 times 10 to the minus three temperature is 2 98 lawn of Q Q Is the concentration of the products divided by the concentration of the reactant CE. So that is 2.4 times 10 to the minus. Three times 1.5 times 10 to the minus six, divided by 0.98 on this will give you a number off. Negative 23 point three Killer Jules.

In each part of this problem were given a balanced chemical reaction in a value of K, C and T. And from this information, we want to find KP for the given reaction. So we're trying to find KP from this equation, and we know values of K, C and T, in order is a constant. So we need to first determine built and gas the change in the total number of of moles of gas between the products and the reactions starting with the given equation and port A. We see that on the product side, we have one gashes species and there are two moles of it. So we have a total of two moles of gas on the product side, minus one mole of this gas. On the reactant side, that means that Delta and gas is equal to one. Now we plug in everything we know into this equation to solve. For KP AP this sequel to K C, which were given a 6.1 times 10 to the negative third. That's 6.1 times 10 to the power of negative. Three times are T, where R is the ideal gas constant 0.0 8 to 1 Leaders times, atmospheres for mole. I'm skilled in he is the temperature given as 298 Kelvin. And now we raise this to the power of delta and gas, which we just found was one. So that means that KP is just equal to K C times rt. When we multiply those three quantities together, we should get an answer for KP to be about 0.15 And now we do the same thing for a different reaction in a different given K c and t for part B, you begin again by finding the delta and gas in on the product side, we have que moles of the gas That's two total moles of caches products minus one mole of the gas plus re moles of the gas for a total of formals of gashes Reactant. And that means that delta and gas comes out to be negative too. Now we just use that equation again to solve for K c. The Eagle to KP. Excuse me. I mean, KP is what we're solving for were given a value for K C, which is 2.4 times 10 to the negative. Third, 2.4 times 10 to the negative. Here it corresponds to that Casey giving a party multiply by rt are again is your 0.8 to 06 leaders times atmospheres or more times killed in temperature Given in part B is 1000 Kelvin and we raises to the power of delta and gas, which is negative. Two. We calculate that out and we should get a final answer or KP be about 3.6 times 10 you the negative seven.

Chapter 16 Problem one fourteen 14. Chemistry's the Central Science. So the question gives us two equations for glass stains for the K and the for the K B when an equally graham with water and the first question access to use the values of K and K B tio this estimate and the equilibrium constant for the inter molecular protein transfer for the word of God. So to do this, we know that our K equilibrium is going to equal to K a times K B. And when every plugging our numbers from that as listed above, you do 4.3 times 10 to the third times 6.0 times 10 to the negative five. And when you play that into your calculator, you get 2.6 times 10 to the negative seven. And so that is going to be your answer for apart, eh? So for part B, it's access. What is the ph of a pointer of by Muller equi? A solution of slicing. And so did any of this. We're gonna have to use an ice table and a K A value to people to get the answer. So the first thing we need June's right out of the equation to me. But like it, this so we're going to do. Hmm? Is we have R H three n c h to CEO Oh, minus in this plus plus, our h two well goes to in each too. Sorry, guys. Goes to each two in CH to CEO minus plus each 30 plus. So now that we have this, we can know Figure out what our eat equation is going to being, and we can use our ice people from there. So we know we have an I c e in our initial concentration. Was there a 0.5? Since we have water that does not play a role here. And also, we don't have scissors very concentration off for and, um, each fio plus. So now what we're going to do is then right, and the exchange is gonna happen. So we're going to subtract something from the initial concentration, and then we're gonna add that to the other side, and then we're gonna have the equilibrium where we add the two scissor cleanser of five Mueller minus x x and then X. So the next thing we have to do is actually find our K a value, especially since we had that h three plus we know is a K and K p and be able to find our cave. Eh? We had to use the equation given above that matches the equation. K W is equal to K a tanz came B where Kay is going to go on a k w divide by K p. And so whenever we plug in our values, we get one times 10 to native 14th divided by six times 10 to the negative five. We get a K A value of 1.6 six times 10 to the negative him. It's one thing to note is that since rk A value is very small, it's going to make it insignificant and impact onto ours. There Pointer of five Mueller solution So we can just get rid of that X to make the math a little bit simpler whenever we're doing finding for it. So we have Rick A is equal to our products over reactant and so we're gonna have X squared over Zahra 0.5 and the X Square comes from multiplying the in each to to siege to Seo and then the water concentrations together. And that is equal to the 1.66 times tend to they give him and then all into yourself for X. So X squared is equal to 1.66 times 10 to the negative. Tin times 0.5 all taking the square root We get the value of X to be 2.88 terms. Tintin ing of six. Now having that 2.88 times Tintin egged the six that is going to be our age plus concentration incense that each plus concentration is important to house people tto find her Pete age. We can then just play this in to our equation to find her pH. So all we have to do is our pH is equal to a negative log of 2.88 times 10 to make six, and we get 5.5 floor and our beer interior for people. And then Part C is just asking us what would be the predominant form of classing in solution with a ph of 13. And what would it be in p h of one? So you're wanting to do here is you want to see what John's going to be the most basic form. Which one's going to be the most citic form. And to do that, you can just look at the charges off them as well as your K K B. So for, um Ph of third team the N H, too. Sorry, The H two in CH two minus C o oh is the most basic form of lice in where the H three and minus C H to cede 00 Each is the most acidic for, and the reason is is because it's more able to donate. Those protons were here. We are donated to protons, and so it's with the most basic form.


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