Which of the following solutions will not form a precipitate when added to 0.10 M NaOH?
(a) 0.10 M MgBr2
(b) 0.10 M NH4Br
(c) 0.10 M FeCl2

Question

Accepted Answer

welcome back everyone in this example, we need to write molecular equations for the below precipitation reactions were told that if no precipitate forms to write no reaction. Now according to the prompt, since we have to write out molecular equations for precipitation reactions, we want to recall that precipitation reactant consists of acquis reactant And they will form at least one solid precipitate product. Now to determine whether our reactant, our acquis we are going to have to recall upon our cell viability rules which is something that we should have memorized. So let's move into example one. We have aluminum, sulfate and sodium phosphate. We want to recall that aluminum sulfate consists of two ions where we have aluminum because it's located in Group one A. On our periodic table. Sorry, Group three A. On our periodic table, we recall that it forms the A. L. Three plus caddy on charge. Whereas sulfate we want to recognize as a poly atomic ion where we have sulfur bonded to four oxygen atoms. And from memory we recognize that it forms a two minus an ion charge. Now because we recall that aluminum based on its position on the periodic table is a metal, whereas sulfate is a poly atomic ion or non metal. This is going to form an ionic compound where we have a transfer of electrons and we would form the A. L. Two in parentheses S. 04 and then we have a subscript of three from aluminum's charge as our compound aluminum sulfate. And recalling upon ourselves ability rules for sulfates, we would recognize that all sulfates are soluble outside of the exceptions of barium, lead, calcium and strontium. And in this case we are bonded to aluminum, which is not one of the exceptions of insoluble sulfates. And so therefore this compound is soluble in water and it therefore gets the Aquarius label. So moving on to our second reactant, we have sodium phosphate in this case. And so we're going to write out the ions. So following the same steps, we recognize that sodium is in group one A of our periodic tables and forms the N. A plus one carry on. Whereas phosphate we recall is a poly atomic ion where we have phosphorus bonded to four oxygen atoms where we from memory recognize that we have a three minus an ion charge. And again, we have a metal bonded to a nonmetal, giving us an ionic compound and a transfer of electrons. So we crossed the charges and we formed the N. A three P 04 compound here. And to determine its phase, we recall our Celje bility rules for phosphates where we should recognize that all phosphates are insoluble except those bonded to sodium potassium or ammonia. And in this case we are bonded to sodium and so therefore we would be a soluble phosphate and we get the Aquarius label. And so now we want to move on to form our products where we should recognize that this is a precipitation reaction. So we're going to combine our opposite charges. And so in this case we would have the combination of these two species with their charges to form a product and that would give us the product, sodium sulfate. So we would form a two and then S. 04 when we cross the charges of these two, this is going to be a sulfate product. So we're calling upon our cell viability rules for sulfates. We recall that all sulfates are soluble except those of barium, lead, calcium and strontium. And in this case we're bonded to again, sodium. And so this would be an example of a soluble sulfate. So we get the Aquarius label here since it's dissolved in water. So moving on to our second product, we combine again are two opposite charges. So we're going to combine our sorry, our aluminum with our phosphate and or an ion. And so we would form our second product where we have a three plus caddy in charge that cancels out our three minus an ion charge. And that would give us a L. P 04 as our product. And we should recognize that this is a phosphate. So we're calling upon our solid ability rules for phosphates. We would recognize that all phosphates are insoluble except those of sodium, potassium and ammonia. And in this case we're bonded to aluminum, which is not one of the exceptions of soluble phosphates. And so therefore this is going to be our solid precipitate product because it's an insoluble phosphate. Our last step is to ensure that this equation is balanced. And so we would place the falling coefficients where we have a three in front of our aluminum sulfate, a two in front of our sodium phosphate, a three in front of our sodium sulfate and then a coefficient of just one as we have for aluminum phosphate. And this would complete our molecular equation as a balanced reaction. So this entire equation is going to be our answer for part one of the prompt. Moving on to example two, we have chromium three nitrate and strontium I died. So writing out our ions as we did before, we're going to begin with chromium three, which tells us according to the roman numerals that we form the chromium three plus caddy in charge, recall that chromium is one of our transition metals on the periodic table. And so it can have multiple charges. But in this case we know it's chromium three plus from its name having three roman numerals. Now we recognize nitrate as a poly atomic ion where we have nitrogen bonded to three oxygen atoms and from memory we recognize it forms a minus one and ion charge. Moving on to writing out our compound our compound when we crossed the charges we form C. R. And then in parentheses N. 03 subscript three for our chromium nitrate chromium three nitrate And recalling upon our sorry ability rules for nitrates, we recognize that all nitrates are soluble regardless of what they're bonded to. And so in this case this is going to be a soluble nitrate. And against the acquis label because it dissolves in water. Moving on to our second reactant, we have strontium iodide. So writing out our ions. We recognize that strontium is located in Group two A. On our periodic table and therefore will form a two plus caddy in charge. And sorry, let's write that smaller so S. Are two plus and I died. We also recognize as being in group seven A. On our periodic table and therefore forms the I minus an ion charge. Now crossing our charges here because yet again we have an ionic compound with a metal strontium and I died. Our nonmetal we would form S. R. I. Two. Now we want to recall our sorry ability rules for I died and we would recognize that strontium is not one of the exceptions of insoluble iodide. And so therefore it's soluble and able to be dissolved in water. So we give it the AQ label. Now we want to go ahead and determine our products. So yet again we want to cross our opposite charges here. So beginning with our nitrate and our strontium, we have the combination of these two charges to form our first product where we will form strontium nitrate. So S. R. N. 03 with a subscript of two. Now again, we recognize our cell viability rules for nitrates is that all nitrates are soluble. And so therefore this gets the Aquarius label as being dissolved in water. Moving on to our second reactant, we have again the combination of our charges from chromium and I died. And so therefore we have our product chromium three I died. So we have C. R. I. Three where we should recall that. So this is a soluble chrome, chromium three I died compound. And so it gets the Aquarius label and because we have all this reactant and products, this is actually not a precipitation reaction, but it's rather going to be a double displacement or double replacement. And so in the lab we wouldn't see an obvious reaction here, we would just see two solutions mixing since we don't really form a precipitate here. So this would be our equation. And all we need to do is bounce it out by adding the following coefficient. So we'll have a three here, we'll have a coefficient of three here, a coefficient of three here, another coefficient of three there and then a coefficient of two here for our balanced reaction. And sorry, let's make a correction. So this coefficient here should be a two. So now moving on to example three, we have copper to nitrate. So we should write out our ions recognize that copper is a transition metal on our periodic table. But from the name given in the prompt we recognize that it will form the copper two plus catalon. As far as nitrate we recall. This is a poly atomic ion where we have nitrogen bonded three oxygen atoms and we memorize that it has a minus one and ion charge. Moving on to our compound here, we would recognize we have a metal and a nonmetal. So crossing their charges, we would form see you and then we have N. 032 where we have our cell ability. Rules for nitrates we recall is that they're all soluble. So this gets the Aquarius label because it dissolves in water. Moving on to our second reactant. Riding out its ions, we have lithium which we recall is located in group one A So it forms a plus one Catalan charge and then phosphate we recall is a poly atomic iron where phosphorus is bonded to four oxygen atoms and from memory we recognize it forms a three minus an ion charge. Now crossing the charges because we have a transfer of electrons in are ionic compound, we would form the L I, three P 04 ionic compounds. And this is going to be an insoluble phosphate based on our recollection of our Celje bility Rules now because we don't have to acquis reactant as we need in our precipitation reactions. This is going to be therefore no reaction. And so for our last two answers, we have this equation balanced for number two and then we have this full equation or rather half of the equation, which gives us no reaction as our answer. So everything highlighted in yellow represents our three final answers as far as our molecular equations and whether the reactions are balanced or even former reaction at all. So if you have any questions, please leave them down below, I hope I was able to help, and I'll see everyone in the next practice video.

Which of the following solutions will not form a precipitate when added to 0.10 M NaOH? (a) 0.10 M MgBr2 (b) 0.10 M NH4Br (c) 0.10 M FeCl2

Verified Solution

Key Concepts

Solubility Rules

Precipitation Reactions

Ion Concentration and Reaction Quotient