Assume that the electrical conductivity of a solution depends on the total concentration of dissolved ions and that you measure the conductivity of three different solutions while carrying out titration procedures: (a) Begin with 1.00 L of 0.100 M KCl, and titrate by adding 0.100 M AgNO₃. (b) Begin with 1.00 L of 0.100 M HF, and titrate by adding 0.100 M KOH. (c) Begin with 1.00 L of 0.100 M BaCl₂, and titrate by adding 0.100 M Na₂SO₄. Which of the following graphs corresponds to which titration?

Video transcript

Hey everyone, we're told to suppose that the electrical conductivity of a solution is directly proportional to the total ions dissolved. And using tight rations procedures, the conductivity of three different solutions were measured, match the following graphs to the correct attrition. So for our first one we have a 0.150 molar sodium hydroxide is added drop wise in a 1.0 liter of 0.150 moller hydrogen cyanide solution. For this question, we need to determine our overall reaction and our ionic equation in order to see the total ions dissolved in our reaction. So starting with our first one, we have hydrogen cyanide and this reacts with sodium hydroxide. When these two react, we get sodium cyanide plus our water. Now for our ionic equation, we have hydrogen cyanide plus our sodium ion plus our hydroxide ion. And we end up with our sodium ion plus our cyanide ion plus our water. Now we know that hydrogen cyanide is a weak acid and does not dissociate completely. So this means our first solution will have the lowest initial conductivity. So looking at our graphs, it looks like C is going to be our first solution. Now let's go ahead and look at our second solution. For a second solution, we have a 0.150 moller calcium chloride is added drop wise in a 1.2 liter of 0.150 molar sodium phosphate solution. So our overall reaction here is going to be our sodium phosphate and this is going to react with our calcium chloride. When these two react, we get our calcium phosphate plus our sodium chloride. Now, to balance this equation out, we need to add a coefficient of two prior to our sodium phosphate. A coefficient of three prior to our calcium chloride, a coefficient of two prior to our calcium phosphate and a coefficient of six prior to our sodium chloride. Now when we write our ionic equation, we have six of our sodium ions Plus two of our phosphate ion, Plus three of our calcium ion Plus six of our Chlorine Ion. Now, in our product side we have two of our calcium phosphate Plus six of our sodium ion Plus six of our Chlorine Ion. Now looking at our ionic equation, we can see that initially we had eight ions and it would be our six of our sodium and two of our phosphate. Now, before we determine which graph matches this solution, let's go ahead and look at solution three. First for solution three, we were told that we had a 0.150 molar sodium bromide is added. Drop wise in a 1.2 liter of 0.150 Moeller led to nitrate solution. So our overall reaction is going to be our lead to nitrate plus our sodium bromide. And we produce our lead to bromide plus our sodium nitrate. Now, to balance out this reaction, we need to add a two prior to sodium bromide and a two prior to sodium nitrate. Now let's go ahead and write our ionic equation. So in our react inside we have our lead ion Plus two of our nitrate ion Plus two of our sodium ion Plus two of our browning ion. And in our product side we produced our lead to bromide, Plus two of our sodium ion, plus two of our nitrate ion. So initially we had three ions. So this means our solution to had the highest initial conductivity While our solution three had the intermediate initial conductivity. Now let's go ahead and match our graphs. So it looks like solution to is going to be graph A. And it looks like Solution three is going to be Graph B. And these are going to be our final answers. Now, I hope this made sense. And let us know if you have any questions.