HF is a weak electrolyte and HBr is a strong electrolyte. Which o...

Question

HF is a weak electrolyte and HBr is a strong electrolyte. Which of the curves in the figure represents the change in the boiling point of an aqueous solution when 1 mole of HF is added to 1 kg of water, and which represents the change when 1 mol of HBr is added?

Accepted Answer

Welcome back, everyone. We have Chloric acid and chloris acid, which are both oxy acids. But chloric acid is a strong electrolyte. While chloris acid is a weak electrolyte identify the curve in the illustration that depicts the change in the solution's boiling point when the solution is made up of one mole of chloric acid and one kg of water. And the change when the solution is made up of one mole of chloric acid and one kg of water. Our given curve below describes a, a graph with actually two curves in which on our y axis, we have vapor pressure expressed in ATM S and we have temperature expressed in degrees Celsius. We can visualize a purple curve here that is given to us in which our purple curve because it is higher than our second curve, which is green. It's going to lie at a higher vapor pressure based on its higher position of our Y axis. And then because it is further to the left of our x axis, we can understand that our purple curve represents a lower temperature in degrees Celsius. Next, we have a green curve which again is lower than our purple curve and further to the right. And so this will correspond to a lower vapor pressure, however, a higher temperature in degrees Celsius. Now, with this outlined, we're going to need to recall our formula to calculate the boiling point change of a solution in which we can take the van Hoff factor of that solution represented by the symbol lowercase I multiplied by the boiling point constant of water. K B multiplied by lower case M which is our modality of our solution. And so recall that modality again expressed by lower case m sorry, this is modality here is calculated by taking our moles of our solute, which is our substance being dissolved divided by our kilograms of solvent. In this case will be water. Now, because we have two curves which represent two different solutions. We're going to say we have solution A as our first solution in which we can assign our first solution one of our acids. So let's begin with chloric acid H C L 03 and recall that chloric acid is a strong acid. It's one of our memorized strong acids from general chemistry. Now recall that strong acids will fully dissociate so that we form as products, one mole of protons H plus and one mole of our chlorate C L +031 minus anion. Now let's recall that four strong acids will typically have a van Hoff factor greater than one. Now, we can look at the complete dissociation of chloric acid. And see that we have one mole of protons and one mole of our chlorate anion, which corresponds to two dissolved ion particles. And so we can estimate that our Van Hoff factor is equal to roughly two. So next, we have our second solution for our second curve which will say a solution B and which we'll assign chloris acid to H C L 02. Recall that chloris acid is one of our weak acids because it's not on our memorized list of strong acids. And so recall that weak acids will only undergo partial dissociation. So when dissolved in water, we will have an equilibrium with the following ion products where we form one mole of protons H plus and one mole of our chloride C L minus and ion. And so because of this partial dissociation, and also because of the fact that we should recall that four weak acids will have a van ho factor which ranges between 1 to 2. However, again, because of the partial dissociation and load dissociation factor, we would therefore estimate that our Van Hoff factor is closer to a value of one. Now, with our Van Hoff factor variables determined we're going to need to determine modality. We're told that for our first solution solution, a we have one mole of our chloric acid per kilogram of our solvent water. So solving four modality, this will give us a modality of one mole for our chloric acid solution, H C L 03. Next, for solution B, we're told that we have one mole of chloris acid, H C L 02 per kilogram of solvent water. And so counting or calculating the modality here, we'll find a modality of one mole of chloris acids of our chloris acid solution. Now, with our modalities calculated, we can clearly see that the modality of solution A is equal to the modality of solution B. And this will make our calculation for boiling point change pretty simple. And so calculating first our boiling point change of our Chloric acid solution H C L 03, we would again take our Van Hoff factor I multiplied by our boiling point constant of water. K B multiplied by lower case M our modality of our chloric acid solution recall that K B is not given to us. And from our textbooks, the boiling point constant of water is equal to a value of 0.512 degrees Celsius per mole. And so plugging in our variables into our formula, we have the van Hoff factor of our Chloric acid solution, which we determine to be a value of two, multiplied by the boiling point constant or K B of water being 20.0.512 degrees Celsius per mole and multiplied by our modality of our Chloric acid solution, which we determine to be one mole. So canceling out our units of Mola, we're left with degrees Celsius as our final unit. And this is going to result in a boiling point change of our chloric acid solution equal to about 1.0 degrees Celsius. And again, this is an approximation because we estimated our Van Hoff factor. Next, let's calculate our boiling point change of our Chloro acid solution H C L 02 in which we would take its van Hoff factor, which we determined to be one since it only partially dissociates multiplied by the boiling point constant of water. K B 0.512 degrees Celsius per mole multiplied by our modality of our chloris acid solution. One mole canceling our units of mola. We're again left with degrees Celsius and we will find an approximation of our boiling point change of our chloris acid solution equal to 10.0.5 degrees Celsius. So as we stated above, our purple curve corresponds to a higher vapor pressure and a lower temperature change. And so based on our boiling point change of our chloric acid solution being lower as 0.5°C. We would assign our Chloro acid solution as the boiling point change being represented by the purple curve. And then based on our notes above for our green curve which lies at a lower vapor pressure, but a higher temperature change in degrees Celsius, our calculated boiling point change of chloric acid will be equal to 1.0 degrees Celsius will be representative of our green curve. And so going back to our given graph, we can label our green curve as our chloric acid solution H C L 03 solution and our purple curve as our chloris acid H C L 02 solution. And so our final answers are going to be that our chloris acid solution corresponds to the purple curve and our chloric acid solution corresponds to the green curve. So this is our final answer. I hope this made sense. And let us know if you have any questions.

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HF is a weak electrolyte and HBr is a strong electrolyte. Which of the curves in the figure represents the change in the boiling point of an aqueous solution when 1 mole of HF is added to 1 kg of water, and which represents the change when 1 mol of HBr is added?

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