In aqueous solution, sodium acetate behaves as a strong
electrolyte, yielding Na+ cations and CH3CO2
- anions. A
particular solution of sodium acetate has a pH of 9.07 and
a density of 1.0085 g/mL. What is the molality of this solution,
and what is its freezing point?

Question

In aqueous solution, sodium acetate behaves as a strong 
electrolyte, yielding Na+ cations and CH3CO2 
- anions. A 
particular solution of sodium acetate has a pH of 9.07 and 
a density of 1.0085 g/mL. What is the molality of this solution, 
and what is its freezing point?

Accepted Answer

Hi everyone for this problem. It reads calculate the morality and freezing point of a solution of sodium beauty annoy. That has a ph of 8.96. And a density of 1.01 g per mil a leader and a quick solutions sodium butin oy acts as a strong electrolyte. Note that the acid dissociation constant for Beauty in OIC acid is 1.48 times 10 to the negative five. So what we want to do here is calculate the morality and freezing point. Okay. Of a solution of sodium Beauty annoyed. So let's go ahead and get started. So the first thing that we're going to want to do is we can solve for the concentration of hydro knee um ions and our concentration of hydro ni um ions is equal to 10 to the negative P. H. Okay. And so this is going to equal 10 to that. We were given the ph in the problem we're told the ph is 8.96. So this becomes 10 to the negative 8.96. So we get a concentration of hydro knee um equal to 1. times 10 to the negative nine. And we can solve for the concentration of hydroxide as well. Because the equation that we can use for that is K. W. Is equal to the concentration of hydro ni um ions times the concentration of hydroxide ions. So if we want to solve for hydroxide or the concentration of hydroxide ions. We're going to take K. W. And divide that by hydro ni um So our KW is a value that we should know and that value is one times 10 to the negative 14 and we just saw for our concentration of hydro knee. Um so we have 1. 64 Times 10 to the -9. And this gives us 9. times 10 to the negative six molar. And this number it will serve as X in our ice table. Okay so well right here serve as X and equilibrium. Okay so we're told in the problem, another piece of information is that the acid dissociation constant for beauty, folic acid is 1.48 times 10 to the negative fifth. What we can do with this is we can solve for the KB which is the based association constant. And the way that we do that is we take K. W. And divided by K. B. So then what we're going to get then is K. W. As we said is one times 10 to the negative 14. And we're going to divide this by K. Eh excuse me, I made an error there. So this should be K. W divided by K. A. And we just saw for the K. A. So let's go ahead and put that value or no K A. Was given. So we have 1.48 times 10 to the negative fifth. So our KB which is our base association constant is equal to 7. 757 Times 10 to the negative 10. Okay, so now what we can do is we can create an ice table and we can use that ice table to solve for our initial concentration. Alright, so we're going to do is we're going to write a reaction and that reaction is going to be between the conjugate base of beauty in OIC acid reacting with water. Okay, so what that's going to look like is Beauty rate reacts with water and this is at equilibrium to produce domoic acid plus hydroxide. Okay. Now with an ice table which is what we're about to create here. So we can go ahead and write ice on the side. We don't include, we don't include solids or liquids. So we're going to ignore this water here for the iro of our ice table. This represents initial concentration. Okay. And we're going to let why represent our initial concentration for beauty rate? And we don't know what this initial concentration is. We need to solve for it and we have zero for both of our products. So what that means then is our reaction is moving in the forward direction and our concentration of the reactant is going to decrease and our concentration of products is going to increase. So our change for the reactant is going to be minus X. You only have one mole of it. So it's just minus X. And for our products it's going to be plus X. So let's go ahead and combine our I and R C ro. And so we get for our react mints at equilibrium y minus X. And then this we bring the X is down. So now we can write our KB expression. Our KB expression is our based association constant expression and it is equal to concentration of products over concentration of reactant. Okay, and looking at our reaction, what this is going to equal then is our products which is the, you know, it's acid and hydroxide over our concentration of reactant. And remember we don't include solids or liquids. So this is just going to be the butyrate. Okay, and what we can do here is we're going to replace what's in the equilibrium row of our ice table into that value. Okay, so then what we're going to get then is we already know what the value is for KBKB is equal to we saw for it is equal to six 756757 times 10 to the negative 10 and this is going to equal So we're going to plug in what's in the equilibrium row of our ice table for these. So we're going to get times the concentration of X times X over the concentration of y minus X. Okay, so with that we're going to replace our X with what we saw for up above. Okay, so remember we said this number here is going to serve as X in the equilibrium. Okay, so we're going to plug that number in. So what we're going to get then is this wrote this numerator here essentially becomes X squared. Ok. And that number that we said will serve as X goes here. So when we simplify this, what that becomes is 6.756757 times 10 to the negative 10 is equal to 9. times 10 to the -6 squared. And this is all over? Why minus 9. times 10 to the negative six. Okay, so now what we're going to do is get rid of our fraction. Okay, So we're going to take we're gonna multiply both sides. So we're gonna take 6.756757 times 10 to the negative 10. And we're going to multiply that by why minus 9. times 10 to the negative six is going to equal 9.120110 times 10 to the negative six squared. So we eliminated that fraction. So now let's go ahead and foil out the left side. And when we foil out the left left side we get 6. times 10 to the negative 10, Y minus 6. times 10 to the - Is going to equal 8. times 10 to the -11. So now we can solve for why that is our goal here. So what we're going to do then is we're gonna add 6.162237 times 10 to the negative 15 to both sides so that we can isolate that Y. And when we do that, what we get is 6. times 10 to the negative 10 Y is equal to 8. 25, 7 times 10 to the -11. Now, why is going to equal 8. 25, 7 times 10 to the -11 divided by 6. 56 75, 7 times 10 to the -10. And what we'll get for Y is Y is equal to 0.123110. And this is going to equal the the initial concentration of the conjugate base. Okay, so let's write that here. And we'll write initial. So this is the initial concentration of the conjugate base. And X. Is going to equal what X. Which we know equals 9. times 10 to the negative six equals the concentration of beauty in OIC acid. Okay. And hydro knee. Um and hydroxide at equilibrium. So let's go ahead and write that. So this is equal to at equilibrium which is also equal to the concentration of hydroxide at equilibrium. Okay, so now what we can do is we can solve for the concentration of the conjugate base at equilibrium because we know what it is initially but we don't know what it is at equilibrium and we can use what we have on the ice table to do that. Okay, so at equilibrium based off our ice table it equals y minus X. And we know what both Y and X are. So let's go ahead and plug this in. So we have 0. minus 9.120110 times to the -6. And this is equal to 0. 101 Molar. Okay, so sodium completely dissociates according to the reaction. So this is what's next. So we have our sodium Beauty No eight and that is going to disassociate according to this. Okay, Okay so what that means then is the concentration is equal to the initial concentration of the conjugate base. So we have our so our concentration of sodium ions is equal to y. Which is equal to 0.123110 moller. Okay so what we're going to do from here then is we're going to assume we have one leader of solution by assuming this. We can calculate the mass of the sodium Beauty annoy salute. Okay, so let's do that. The mass of the sodium Butte. No, eight salute is equal to. So that this number that we have here, The concentration is most per leader. So let's write it out that way. So 0. moles per leader. And we want to go from most per leader of sodium Beauty. No, 82 g. Okay, so to go from moles to grams we need the molar mass. So in one mole they're of this of the sodium, you know it Salt or Yes, a sodium, you know, eight. The molar mass is 100 and 10.88 g. Okay, and now we want to go from g. two leaders. And so we assumed that we had one leader of solution. So we're just going to multiply this by one leader. Okay, so when we do this, the units canceled for moles and the units cancel for leaders and will be left with the units of grams. Okay, so our mass Is equal to 13. g of this sodium. Butin. Oy. Okay, so now we need to know what is the mass of the solution? The mass of solution is equal to mL times the density. So we have 1.1 g per milliliter. Okay, so our middle leaders cancel And we're left with g. So our mass of the solution is 1010 g. The mass of the solvent is equal to the mass of solution minus the mass of solute. So let's go ahead and solve that. So our mass of the solution We just said is 1010 g minus the mass of the solute, which is 13. 934 g. gives us 996. 7066 g. And let's convert that to kilograms by dividing by 1000 and so what we'll get is 0. kilograms. Okay, so now we can solve for morality. Okay, morality is represented by lower case m and morality is equal to the moles of solute over kilogram of solvent. Okay, so our mole of saul you which is our sodium we said is zero . moles. And our mass of the solvent is 0. kg. Okay, So when we saw for this, the answer that we're going to get is 0. 549 um allow. Okay, so morality Is equal to 0. Malala. So that's the first part. Okay, now, the second part of the question asks us for the freezing point of the solution. Okay, so for the freezing point to calculate this, the equation that we're going to use is the change in temperature is equal to I where I is the van Hoff's factor. And for strong electrolytes, i is equal to the number of ions times K. F. Where K. F. Is the morale freezing point depression constant times M where M is the morality of the solution. So let's go ahead and plug in what we need to plug in. So for us to find the van Hoff's factor, we need to figure out the number of ions. Okay, so for our sodium view, annoyed when we disassociate this into its ions, it dissociates into the following. So that means our ions is too. So we'll plug into here and the Malala Freezing Point Depression constant is negative. 1.86°C, Permal al. And our morality, which we just calculated is 0.123549. Um allow. And so what we're going to get then for a change in temperature for the freezing point is equal to 0.459602°C. So now let's remember that the change and temperature for the freezing point is equal to the solution, The final temperature of the solution minus the final temperature of the solvent. Okay, so we know what the change in temperature of the solution is now or the we know what the change in temperature Of the freezing point is. We just saw for that it is negative 0.459602°C is equal to the solution. The change in temperature Of the Solution -0°C. Okay, so that means then that the solution freezing point Is equal to negative 0.459602°C degrees Celsius. So our final answer then is that the freezing point of the solution Is equal to negative 0. 602°C. And that is going to be our final answer for the freezing point of the solution. That is it for this problem? I hope this was helpful.

In aqueous solution, sodium acetate behaves as a strong electrolyte, yielding Na+ cations and CH3CO2 - anions. A particular solution of sodium acetate has a pH of 9.07 and a density of 1.0085 g/mL. What is the molality of this solution, and what is its freezing point?

Verified Solution

Key Concepts

Electrolytes

pH and Acid-Base Chemistry

Colligative Properties