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TEST FOR Agl+ Remember, vou dissolved the Pb2* ion in hot water and decanted it off; The precipitates remaining in the test tube are solid AgCl and solid HgzClz Hgz...

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TEST FOR Agl+ Remember, vou dissolved the Pb2* ion in hot water and decanted it off; The precipitates remaining in the test tube are solid AgCl and solid HgzClz Hgz and Ag cations can be separated by adding M NHj (aq) to the solid mixture of the two chlorides_ Of those two compounds; only the silver chloride is soluble in aqueous ammonia (NHAOH or NH;): It forms colourless soluble complex ion of Ag(NHzhCl (qk, whereas mercuryll) chloride (HgzClz) turns into Hg metal, Hg and HgNH_Cl (s) which is

TEST FOR Agl+ Remember, vou dissolved the Pb2* ion in hot water and decanted it off; The precipitates remaining in the test tube are solid AgCl and solid HgzClz Hgz and Ag cations can be separated by adding M NHj (aq) to the solid mixture of the two chlorides_ Of those two compounds; only the silver chloride is soluble in aqueous ammonia (NHAOH or NH;): It forms colourless soluble complex ion of Ag(NHzhCl (qk, whereas mercuryll) chloride (HgzClz) turns into Hg metal, Hg and HgNH_Cl (s) which is greyish-black precipitate: HgNH_Cl is insoluble in ammonia_ AgCl (s, white) 2 NHz (aq) = [Ag(NHsh] : (aq) + Cl (aql (This net ionic equation ] HgzClz (s, white) 2 NH; (aq) = Hg black) HgNHzCl (s, white) NHCl (aq) [This net ionic equation ] The precipitate will appear to be grey but actually is mixture of black, Hg white, HgNHzCl (s} PART To the test tube containing vour white silver and mercury precipitates, add drops of 16 M (concentrated) NHAOH (NH;): Caution: concentrated ammonia is very irritating to the nose The mixture in the test tube will form soluble complex ion of Ag(NHs)Cl (aal and an insoluble Hg metal, Hg (IJ and HgNH,CI () complex: It will appear as grevish-black precipitate This Is positive test for HBz ion. Centrifuge the contents of the test tube: Save the supernatant solution since it now contains silver in the form of the aqueous Ag(NHslz ion. The solid mercury compounds will remain in the test tube; AgCl (s, white) 2 NH; (aq) = [Ag(NHz)z] 1* (aq) (aql [This Is net ionic equation:] PART Il: To the test tube containing the aqueous silver complex, [Ag(NHshz] 1* Cl 1 (4q/ (supernatant)] add drops of 16 M HNO; (Caution this is very strong and nasty acidl) The solution should be acidic at this point: The acidity can be tested by dipping stirring rod into the solution and then touching (with drop of solution) to piece of blue litmus paper. If the solution is acidic it will turn the blue litmus paper red. Red litmus paper stays red in the presence of an acid; The formation white precipitate of AgCI in the acidic solution is confirmation test for Ag] The soluble diamminesilver(I) ion of [Ag(NH;)z] (aqk will react with the acid and precipitate out as white solid AgCI This is positive test for the Ag ion_ [Ag(NH;h] " colourless| (aa) 2 Hl' (aq) = Agci white) 2 NHa (aql [This net ionic equation:] [Ag(NHshz] (aa) +2 Hl" (aq) AgCl () + 2 NH;] (aql (This is . net ionic equation:]



Answers

Part 1: (a)Ammonia, $\mathrm{NH}_{3},$ is a weak electrolyte. It forms $10 \mathrm{~ns} \mathrm{n}$ solution by reacting with water molecules to form the ammonium ion and hydroxide ion. Write the balanced chemical reaction for this process, including state symbols. (b)From everyday experience you are probably aware that table sugar (sucrose), $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11},$ is soluble in water. When sucrose dissolves in water, it doesn't form ions through any reaction with water. It just dissolves without forming ions, so it is a nonelectrolyte. Write the chemical equation for the dissolving of sucrose in water. (c)Both $\mathrm{NH}_{3}$ and $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$ are soluble molecular compounds, yet they behave differently in aqueous solution. Briefly explain why one is a weak electrolyte and the other is a nonelectrolyte. (d)Hydrochloric acid, HCl, is a molecular compound that is a strong electrolyte. Write the chemical reaction of $\mathrm{HCl}$ with water (e)Compare the ammonia reaction with that of hydrochloric acid. Why are both of these substances considered electrolytes? (f)Explain why HCl is a strong electrolyte and ammonia is a weak electrolvte (g)Classify each of the following substances as either ionic or molecular.
- $\mathrm{KCl}$ $\mathrm{HCl}$ - $\mathrm{NH}_{3}$ $\cdot \mathrm{Ca}(\mathrm{OH})_{2}$ $\cdot \mathrm{CO}_{2}$ $\bullet \quad \mathrm{PbS}$ $\bullet \mathrm{MgBr}_{2}$ $\bullet \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ (h)For those compounds above that you classified as ionic, use the solubility rules to determine which are soluble. (i)The majority of ionic substances are solids at room temperature. Describe what you would observe if you placed a soluble ionic compound and an insoluble ionic compound in separate beakers of water. (j)Write the chemical equation(s), including state symbols, for what happens when each soluble ionic compound that you identified above is placed in water. Are these substances reacting with water when they are added to water? (k)How would you classify the soluble ionic compounds: strong electrolyte, weak electrolyte, or nonelectrolyte? Explain your answer. (l)Sodium chloride, $\mathrm{NaCl}$, is a strong electrolyte, as is hydroiodic acid, HI. Write the chemical equations for what happens when these substances are added to water. (m)Are $\mathrm{NaCl}$ and HI strong electrolytes because they have similar behavior in aqueous solution? If not, describe, using words and equations, the different chemical process that takes place in each case. Part 2: You have two hypothetical molecular compounds, $\mathrm{AX}$ and AY. AX is a strong electrolyte and $\mathrm{AY}$ is a weak electrolyte. The compounds undergo the following chemical reactions when added to water. $$ \begin{array}{l} \mathrm{AX}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{AH}_{2} \mathrm{O}^{+}(a q)+\mathrm{X}^{-}(a q) \\ \mathrm{AY}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{AH}_{2} \mathrm{O}^{+}(a q)+\mathrm{Y}^{-}(a q) \end{array} $$ (a) Explain how the relative amounts of $\mathrm{AX}(a q)$ and AY $(a q)$ would compare if you had a beaker of water with $\mathrm{AX}$ and a beaker of water with $\mathrm{AY}$. (b) How would the relative amounts of $\mathrm{X}^{-}(a q)$ and $\mathrm{Y}^{-}(a q)$ in the two beakers compare? Be sure to explain your answer.

So what we're looking for here is balancing equations where we have the same number and type of atoms on each side of the equation and the faster we have three see you at two and three at a H plus. That is an equilibrium with three. See you two Plus that two N o God for H two. Next. What we have is to see you two plus up to S C N at H s. Three h 20 has an equilibrium with to see U. S. C n hs four. Add to H plus. Next we have 10. See you plus to I owe 3 12 h plus 10. See you two plus Got make delivery tomorrow I two at six. H 20 in the next one. What we have, I too have to us. 2032 miners. That is an equilibrium with two i miners, but ask for six tu minus. Finally, what we have is to said on N H four p 04 at c n. That is an equilibrium with CNN P two 07 at H 20 and each three where we have two of those. So in the second part. The amount of copper there's not 0.42 g on percentage wise that it's 77.2% mama's off C N two p, 20734.6 The amount of sink There's not 0.1055 grounds on the Mass. 19.4%.

So I am. Exchange chromatography is executed by making the reversible exchange of irons between irons present and solution on iron exchange resin. So we tend to use this to separate similar irons. And so in the first part where it's calculating the total amount of solution fed into the column 36 7.2 liters. So in part B, the expression of a volumetric flow rate for solution a be a T think was not point for liters per minute. The VA 33.6 minutes multiplied by t was able to not point not 11 90 so we can apply the mass balance in solution A M put a is equal to accumulation at a and we are able thio. The salmon that has Time T varies from not to 33.6 minutes and V A varies from zero to V a bracket. L The volume of Solution A becomes 6.72 liters on. We can calculate the amount of sodium chloride there's an A. C. L is equal to 392.72 grams and a c l. And so we can give an expression for environment for a lower rate of solution be that is tu minus no point no. 11 19 that is equal to be be dark t. So the total volume of solution a 60.48 liters total volume solution by solution be not at all number of moles of tres fad. Three points. There are two full mall on some possible potentials for error. So pumped be might have failed, causing a flow of one mole of sodium chloride to the column during the illusion period. Programs method chosen to execute in the chromatography could be incorrect. Perhaps the column was packed with the wrong reason, maybe arising. That has reduced affinity compared to the correct resin.

So between the lithium night treat, the nickel to night tree and the strong team nitrate, the one that we should actually used to precipitate the maximum out of the carbonate ions from the solution should be the strong team nitrate. And the reason for this is has the lowest ks beef value asking here. And so they gave us a case be table with the lithium carbonate, nickel to carbon and the strontium carbonate. And so adding strong human nitrate trying him carbonate is actually what we'd be making. So it would make sense for look at these case p values. And so when a case be value is closer to one, that means that the, um substance dissolves very readily and be squeak. It's only ions and it doesn't like to stay in the forms of for instance, lithium carbonate. The case be was one then what we would have is only the theme ions and only the carbonate ions. We would actually have the lithium carbonate, but here so lithium carbonate has the value closest Thorne and strong from carbonate has value closest zero, which means that it likes the state s surround him carbonate and so once we put the strontium nitrate into the solution and it reacts with the carbon e ions, we get the strong carbonate and have a lot of precipitated. Then for one beer, what we're looking at is for the carbon compound that contains the cattle on chosen in part a, which we chose as strontium nitrate, giving us drunken carbonate determine the concentration of each eye on of that compounded solution equilibrium. So this is the equilibrium, expression or equation. And so what? We have your strong him carbonate solid, solid in equal air be in, which is what these double arrows are with strong team two. Plus, it's the cat eye on and it's a quest. And then the carbon, eh? I unders and own. And that's also Equus. So then, um, what we need now is chaos P is actually just an equation, like it's just a formula that you follow. So they gave us the KSB and Cass be equals the concentration of whatever is a quest on the right side of your equation. So we have the strong tea, um, cat eye on multiplied by the concentration of the carbon A an eye on. So now since they gave us our case, P value, which is just 5.60 times 10 to the negative 10th. We're setting that equal to basically just the concentration of the strong Thiam times, the concentration of the carbonate and what we can do there is that's just x Times X giving us X squared. So now if we just take the square to both sides and we saw for eggs X equals the concentration of the strontium I on which is also equal. So the concentration of the carbonate ion, which overall is equal to 2.37 times 10 to the negative fifth Moeller. And so that's the concentration off each eye on in solution. So now moving on part C one mixing the solution should the student ensure the carbonate solution or the nitrate solution is an access. So for this, what we have to think about it, Which one of these ions would actually help us more? So we are Actual precipitated is strong team carbonated, which is as our CEO three. So we don't need more carbon ian solution because it doesn't matter how much carbonate we have, because that's just gonna stay as an eye on but for the strong tea Um, the way that we got the strong team is from the strong tea of nitrate, which is this So if we don't have enough strontium, then we can't make the strong him carbonate. So the nitrate solution should be an access because in order to create the maximum amount of precipitate, you need enough strong team I owns to react with all of the carbonate ions and having the access strontium ions in the solution doesn't affect the master of the carbonate in the virginal example. So we have a fixed amount of carbonate, but we I don't know, having more having the access won't give us the precipitate but having the excess of strong here Mayan means that we can react all off the carbon e into precipitously and then having the extras drawn to him doesn't affect anything. So then moving on to part dear after titrate and sufficient solution to precipitate out all of the carbonate ions, the student filters the solution before placing in the crucible and heating it to drive off the water after several beatings. The final mass of the precipitate it remains constant is determined be 2.2 grams. So the first thing we wanted to do is determine the number of moles of Pacific it. So what we d'oh is we take our 2.2 grams of our strong team carbonate, and we multiply that by the molar mass, which is basically just a ratio that allows us to go from grand symbols or moral stra graham. So 1 47.63 grams Uh, s or C 03 her one more and overall, that gives us okay after I put it into my calculator one point 37 times tend to the negative seven Negative, second more. And now, um, moving on determining the mass of the carbonate present in the precipitate. What we do is we take these moves that we just got and then we multiply that bye. You have one more of, um, you have one mole of the carbon e ion her one more of the strong him carbonate substance. So then, since it's just a wonder one we end up with 1.37 times tend to the negative second more of the carbonate and then what we need to do is take the Moler Mass off the carbon e ion and multiply that by the moles. And so once the molar mass of carbon e is so carbon is 12 plus the 40 the 48 from the three oxygen. So it's 60. So then once we take 60 times so 0.137 times succeed, which gives us zero point e to two grams of this carbon ee eye on. So then, um, poor E. What we need to do is determine the percent by massive carbonate in the original sample. So the 0.822 grams of the carbonate weakness around that 20.82 grams and just carry the 26 eggs and then you divide that by don't know why I did that. Um, so it's the zero point 82 grams divided by the one point 39 grams from the original sample. And then if you multiply it by 100 that's how you're gonna percent and we end up with 43.4 percent of the carbon E who of the, um carbon e i aunt in the original solution so that moving on too. Oh, you know, I did that. So then moving on to part, as is the original compound most likely lithium carbonate, which is ally to Theo. Three, um, sodium carbonate ing any to Theo three or potassium carbonate, which is you if you go through here. So what we do now is so it's the, um 60.1 for all three, um, on the numerator because what we're looking at for all of them is the carbonate. But then the only difference is the total math. So we're either dividing it by 73.89 because you had the mass of the carbonate plus the mass of dilithium or 16.1 divided by one of 5.99 or 60.1 divided by 1 38.21 Oh, and so you multiply all these by 100 and you enough with anyone 0.2% for that 56.6% for this and 45 0.4% for the pizza, potassium carbonate and

Problem 94 party. Yeah, silver chloride solid dissociates into silver echoes plus serial negative. Of course. This is our answer. Apart 1st, in part of the So Casey equals two concentration of silver and multiply by concentration of chloride in because we can't consider as We can assume the concentration of solid. It's one this answer apart B in part C. Using thermodynamic data from appendix calculate delta as for reaction in the party so we can get through here. It's not goes to very bad. It's not frozen of silver saluda in khost medium plus delta. It's not He was an up chloride iron in khowst media minus delta. It's pure reason in solid. A test version of silver right and solid. So after putting the value There it adds not equals two. Putting the value 10 hi 0.9 zero kg jule minus one 67 .2 kg jule minus -127 0.0 kg june That is equals two 65 point seven. You go to okay to this positive sign. Mhm. Project. So this positive sign indicate that that reaction is in the thermic in long term it so celebrity of at this year's solid in as to liquid will increase with increasing the temperature so soluble itty of silver chloride solid in ads to liquid will increase. Story will increases with increasing temperature. This is a for answer up part C. In part of the calculate the first we need to calculate the celebrity of silver chloride solid in pure water at silver Ryan in or end in 0.100 M sodium chloride echoes etcetera. And in zero point 100 am sodium chloride echoes as silver chloride irons so can solve it. Okay your water in pure water leg concentration up sell your iron calls to concentration of chloride iron. That is because two x. So Casey Casey Equals two silver iron concentration Selena and into concentration of chloride iron. That is because to access choir That is equal to 1.6 into 10. To the power minus 10 so X equals two. 1.6 Into 10. to the power -10 to the power one way to so concentration of steel or iron in pure water. That is equals two. One point three into 10. To the power-. Hi I am okay and concentration of silver I am in 0.100 am sodium chloride equals a dooming. No poor medicine. No formation up silver at the two cl two negative so concentration of steel or iron mm 0.10 am sodium chloride is because two X. Therefore concentration of C. Alliance See all negative and equals two zero point 100 bless X. Because uh 0.0.100 chloride and already presenting a solution. Therefore case equals two concentration of silver and multiply the concentration of chloride time. That is because two X. Multiply way 0.100 less X. That is equals two 1.16-10 to the power minus stand after solving. Sorry, Assuming the X. is small relative to 0.100. The airport weekend, ignore this. Ex this ex therefore zero point 100 X equals two 1.6 into going to depart ministry. Okay so X equals two 1.6 into 10 to the Power -9. The concentration of silver Ryan In G 0.100 Amsterdam correct? Sorry not this is our answer. Okay mm account for complex complexes and the formation of salute, salute at D. C. L. Two minus some the two reactions and calculate K. for overall process. So first we right silver chloride solid the associated into see lower iron in Pecos medium plus chloride. Iron in equals million. Okay and let's give her this reaction is 1.6 into 10. To the power understand given the question and silver in echoes medium. Bless to see a negative in echoes medium that will give at the Cl two negative in a post medium. And this In this reaction can let's reaction intent is K two and K two. We can You too. For this reaction is one .8 into 10 to depart right after adding these two reactions you get silver chloride solid plus where I am echoes. Yes to seal negative I am echoes was to silver iron echoes plus chloride. Iron echoes plus At the cl two negative equals and Casey for This reaction equals two. Even multiplied by kid too. Given multiple advocate to So after simply fine. We clearly see here this one this one clothing and this one clothing is cancel out this solution this seller and this slower is also cancel out. So we can write. Yeah actually serial solid less. See a negative across it. Will the associated sorry. Associated to form at the two the L two minus equals. Okay and you see for this reaction isn't multiply here and we get 2.9 into Came to the Power -5. Okay body baby? S. Question. Okay so we can write Casey for Casey for this reaction. So Case equals two concentration of product and considers not reacting. So concentration of entity cl two. Sorry this is also not too Okay divided by concentration of chloride in therefore concentration of entity the L two minus minus it equals two X. And concentration of the negative Cause two x. divided by zero point ah 100 minus mexican once. So something not. This is stupid, correct? So concentrate um at this year two equals 2 x. And concentration of chlorine. Hi and it equals two zero point 100 -1. We can already please see here concentration up this. So concentration employed and is here. Take care. Okay Assuming X. is small relative to zero 100 The airport we can write as X equals 2-9 into 10. to the power minus six M That's easy. He'll do we can simply put this and this value in this question. We get this result. Okay will you up this and this and this. So this is an export. This is this is export concentration, this is X. Okay these calculations so that we cannot go in details because you confused So that the cell ability lady of silver chloride, the concentration of any soluble silvery species is actually greatest in pure water. Pure water That is 1.3 in two. 10 to the power minus private. You can clearly see here these are concentration of steel or iron, pure water. The concentration of silver and other medium which is smaller and there's a concentration of silver and other mediums. This is also is moderate. Okay, so in zero point 100 am sodium chloride complex complex. And because complex and cows, the scalability of silver is 2.9 in 20 the power minus six M. Into at at this year. Too negative to be greater than that predicted by common kind of Yeah, Greater than that predicted by the common iron effect alone. That is 1.6 into 24 -9 M. Silver but not as great as its celebrity in pure water. Okay, This explanation, this last part is something tough. So you can carefully read this question. Okay, this is our finances.


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