(b) Calculate the concentrations of all ions in the solution afte...

Question

(b) Calculate the concentrations of all ions in the solution after reaction. Check your concentrations to make sure that the solution is electrically neutral.

Accepted Answer

welcome back everyone. 25 mL of 250.275 molar sodium bromate is added to a mL solution containing 1500.175 molar potassium bromide and 0.4 50 moller cleric acid. The bromate and bromide ions react to produce bromine liquid. After the reaction. What are the concentrations of all the ions in the solution and make sure that the solution is electrically neutral? By checking your concentrations. Let's begin by writing out that reaction between our bromate and bromide ion, which we recognize bromine is located in group seven A on the periodic table, so it forms a minus one anti on charge. And as a product we form according to the prompt bromine liquid. So our next step is to write out half reactions using our re agents. So for our first re agent we have bromate which is going to be reacting with or sorry, it's going to be forming our booming liquid product. And then we have our bromide an ion which also forms brimming liquid as a product. Our first step is to balance non hydrogen and non oxygen atoms first. So looking at our bro mean, we have two moles of bromine on the product side in our first reaction and we only have one on the reactant side. So we'll fix that by placing a coefficient of two in front of our bromate. Now we've introduced a total of six moles of oxygen on the reactant side. And we want to recall that we balance oxygen using water Meaning that we can add six moles of water to our product side so that our oxygen is balanced. But now that we've added six moles of water, we've introduced 12 moles of hydrogen to the product side, meaning we need hydrogen on the reactant side, specifically 12 moles. And so we're going to recall that we can balance hydrogen Using our hydride Canyon. So we would add 12 moles of hydride to the reactant side. And so now we have our atoms balanced in the first reaction. And now for the second reaction we see we have two moles of roaming on the product side. So we'll just add a coefficient of two in front of bromide on the reactant side. So now we have all of our atoms balanced and now we need our charges balanced. So we have a net charge on the reactant side of -10. Or sorry plus 10 on the reactant side because we would have 12 -2 And we have a net charge of zero on the product side. So we need to cancel out that charge of plus 10 on the reactant side here by expanding our reactant side and adding electrons. So that's going to give us a neutral charge on both sides of this of this reaction. Where now in our second reaction we see we have a net charge of minus two. So we're going to bounce that out by adding two electrons to the product side. In our second reaction. Our next step is to bounce out the electrons in both reactions. So we can take two electrons and get to 10 electrons by taking this entire equation And multiplying it by a factor of five. So this is now going to give us 10 moles of our bromide, an ion which forms Five moles of bromine liquid plus 10 electrons. And now we can add this to our equation above where we have two moles of bromate Plus 12 moles of hydride Plus our 10 electrons yields are brimming liquid Plus six moles of water and so canceling out our electrons on the product side in the first reaction and on the reacting side in the second reaction will carry down everything else for our entire balanced redox reaction here, which will give us a balanced equation. That says we have two moles of our bromate, an ion Reacting with moles of hydride Plus our moles of our bromide. An ion Yields six moles of liquid grooming. Since we would combine the five brahmins from the first equation with the one mole of coming from the second equation Plus six moles of water. Now that we have this bounced redox reaction, we need to refer back to the prompt to get our total volume of the solution. So that's our next step. And according to the prompt, we have 25 ml of sodium bromate added to our solution with our potassium bromide and chlorate acid. So we have 25 ml plus 150 ml To give us a total solution of 175 ml as the volume. Now we want this to be in units of leaders. Since we need to get from our concentrations of our re agents in the solution which are given in the prompt in terms of polarity. And we call that polarity can be interpreted as moles per leader. And that is why we need our volume and leaders. So we're going to multiply to convert from middle leaders to leaders By recalling that are prefix Milli tells us we have 10 to the negative third power of our base unit leaders. So we can now cancel out male leaders were left with leaders for volume. And this is going to give us a total volume of . leaders for the solution. And now we can use this to find the initial moles of our ions on the reactant side here. So we're going to begin with our initial moles of our sodium bromate which is our salt Of our bromate ion. And we're going to use that volume that we just found. So .1750 L. We're going to multiply by the molar itty for sodium bromate which is given as .275 moller which were interpreting in moles per liter. And this is part leader of our sodium bromate salt. So you can see we can now cancel out our leaders and we now need to go from moles of our sodium bromate, the numerator, two moles of our sodium in the denominator To get two moles of our bromate ion in the numerator. And we can see in our formula for sodium bromate that we just have one mole of our bromate ion. So we have a 1-1 molar ratio. So now we can cancel out moles of our salt sodium bromate, leaving us with moles of our bromate ion and we have an initial value. Far moles equal 0.0481- moles of our roommate. And I am so following the same steps to get to the molds of the rest of our ions, we begin with our total volume of our solution and multiply by the molar concentration of our salt, potassium bromide. Which is given in the prompt as 0.175 moller which were interpreting in units of moles per liter of potassium bromide. And so canceling out our units of leaders were now going to go from the molar ratio between moles of potassium bromide, two moles of our bromide an ion. And we can see in our formula that we have a 1 to 1 molar ratio because we just have one mole of potassium bromide here. And so now we can cancel out moles of potassium bromide. We're left with moles of our bromide An ion. And this gives us a value of .018375 moles of our bromide an ion. And now we just need our third an ion or ion which is our hydride initial moles. And we're going to begin again with that volume 0.1750 liters multiplied by the concentration of chlorine acid given in the prompt in terms of 0. moles of cleric acid per leader of caloric acid. And multiplying this by the ratio between moles of chloride acid and moles of hydride. We see that we have a again, 1-1 molar ratio. So we have 1/1 here and canceling out our units. We can get rid of leaders as well as molds of clerk acid leaving us with molds of hydride. And this gives us an initial moles of 0. 75 moles of hydride. Now we need to determine out of each of our ions which would be the limiting reactant ion. So we're going to use one of our products being our six moles of bromine to base that off of. So determining our limiting reactant, we're going to begin with our moles of bromate where we have .0048125 moles of bromate. And we are going to multiply this by the ratio between moles of bromate and moles of our products remain liquid where from our bounce reaction above we have a 2-6 Molar ratio. We have six moles of grooming for two moles of bromate. So canceling out moles of bromate. We have moles of our product and this produces zero 144375 moles of our bro me. For our second reactant, we have 0.18375 moles of our bromide. An ion multiplied by the ratio between moles of our bromide, an ion and moles of our blooming liquid, which from our bounced reaction, we see we have a 10 to 6 molar ratio. So 10 moles of bromide, an ion to six moles of grooming liquid, canceling out molds of bromide. We have moles of our product. This is going to yield 0.18375 moles of roaming liquid Produced. And next we have 0.07875 moles of hydride as we calculated above, multiplied by the ratio between moles of hydride and moles of romaine, Which from our balanced equation we see as a 12-6 molar ratio. So we have 12 moles of hydride for six moles of blooming liquid, canceling out moles of hydrogen. We're left with moles of grooming. This is going to yield 0.039375 moles of our roaming that is produced from this region. And we can see that out of each of these amounts of our product produced. The smallest amount of roaming produced is .018375 moles of roaming made from our bromide an ion. And so we would say that therefore our bromide an ion is the limiting re agent. And now knowing are limiting reagent, we can use that information to form an ice chart with our initial changed and final concentrations of each of our ions. So writing out our overall reaction, we have again, two moles of bromate Plus our moles of hydride Plus our 10 moles of bromide produces six moles of burning liquid Plus six moles of liquid water. Recall that in these charts? We don't consider our liquids, we only consider Aquarius re agents. So we have our initial concentrations which we just found for bromate to be Above. We stated that it's 0. most for hydride. We stated it's 0.07875 moles and four bromide. We set at 0.018375 moles. And apologies here, I just caught an error. So for our moles of bromide, we need to correct this and this value here. So that here we have initial moles of bromide equal to 0.0 moles of bromide. Then we take that value here again, we still have a correct amount of our blooming liquid produced here. And then for our ice chart we carry down 0.30625 moles of bromide initially. And so now from our bounced reaction, we will have the change b minus the number of moles we have as our coefficient so minus two X. Here minus 12 X. Here and then minus 10 X here and now for our final concentrations, we're going to figure that out. Since we know that roman is our limiting re agent, we're going to say that it's going to change 10 X by exactly its initial amount. So it's equal to 0.30625. And so solving for X, we're dividing both sides by the right hand integer 0.30625 on both sides to say that X is equal to a value of 0.30625. And plugging that X value in. We'll have our final concentrations so that we have .42 or sorry .042 moles of our bromate Finally and then 0.042 moles of our hydride in the final solution. And then we would have zero of our limiting since it's used up and fully consumed. Now that we have our final concentrations, we're going to use this to verify our electric neutrality but before doing so, we need to find our moles of our sodium potassium and chlorine Or chlorate an ion. So we're going to begin with the total volume of our solution which earlier we determined is .1750 L. We're going to multiply by the concentration of sodium bromate given in the prompt as .275 moller which were interpreting in units of moles per liter of sodium bromate. And sorry, this is not an eye on yet and because this is our salt, so this is then multiplied by the ratio between moles of sodium bromate two moles of our sodium cat ion, Which we can see in the formula is a 1-1 molar ratio. And this is going to result in a value of 0.048125 moles of our sodium catalon. And this is after we cancel out leaders and moles of sodium bromate. Next we begin with our volume of our solution multiplied by the concentration given for potassium bromide being 0. polarity in the prompt which were interpreting in units of moles per liter of potassium bromide. Multiplying then by our ratio between moles of potassium bromide and moles of our potassium caddy on K plus one, Which we can see in the formula is a 1-1 molar ratio which results in a value of 0.030625 moles of potassium. So canceling out leaders and moles of potassium bromide and moving on to our next step which is for our third eye on being our chlorate an ion. We have 0.1750 liters for our volume of our solution. Multiplied by the polarity of cleric acid given as 0.4 polarity in the prompt which we're interpreting as moles per liter of cleric acid and multiplying by the ratio between moles of chloride acid and moles of our chlorate an ion. So that's c l 03 minus which we can see in the formula is a to 1 molar ratio, canceling out our units. We're left with moles of our chlorate, an ion equal to 0.07875 moles of chlorate. And so finally getting to checking for charge neutrality, we're going to begin by finding our total moles of positive charge or catty in charge where we'll take our moles of sodium and potassium. So sodium we have 0.48125 moles of sodium catalon plus r 0.30625 moles of potassium catalon and adding these together as well as our molds of Our cattle on from our ice chart which has a final concentration of .0.042 for hydride. So we add .0.042 moles of hydride here. This is going to result in a sum equal to 0.12075. And now for our total moles of an ion charge negative charge, we have our moles of our an ion being 0.7875 moles of chlorate, cielo three minus plus the moles of our an ion. From our chart over here where we have our bromate an ion which is 30.42 moles at the end of the solution. So 0.42 moles of B r minus this sum equals a value of 0.12075 as well for our total modes of negative charge. And so here we can confirm that our charges balance and so we do have electric neutrality in the solution after the reaction occurs. And sorry this says electric neutrality. Now that we've verified this step, our next step is to finally get to our final answers by getting the concentrations of each of our ions. So beginning with our concentration of sodium were calculated earlier that we have 0.048125 moles of our sodium catalon. And again for more clarity we divide by leaders. So we're dividing by the volume of the solution, which the total volume you calculated is .1750 leaders. And this gives us our polarity equal 2.275 molar of our sodium catalon. And now moving on to our next carryin above, we stated we have 0. moles of our sodium or sorry potassium carryin divided by the leaders of our solution being .1750 l Equal to our polarity of potassium being . molar for our potassium carry on next we have our chlorate and ion which we stated earlier has a amount of moles equal to 0.7875 moles of chlorate cielo three minus, divided by the total volume of our solution. 0.1750 liters. This gives us our molar itty forklore eight equal to 0.450 moller of chlorate. Next we have our boring an ion which we said earlier has a final final moller amount of 0.42 moles, dividing this by the volume of our solution. 0.1750 liters for bromate. So B r 03 minus. We have a concentration of bromate equal to 0. moller of our bromate an ion. And for our last carry on we have our moles of hydride which we determined has a final amount of 0. moles after the reaction of hydride divided by our volume of our solution 0.1750 liters this is going to equal its concentration as 0.240 moller as well for our hydride caddy on. And so for our final answers we've determined successfully the molar concentrations of all the ions in our solution containing sodium bromate, potassium bromide and chlorate acid. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video

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