A 7.0 mass % solution of H3PO4 in water has a density of
1.0353 g/mL. Calculate the pH and the molar concentrations
of all species present (H3PO4, H2PO4-, PO43-, H3O+ ,
and OH-) in the solution. Values of equilibrium constants
are listed in Appendix C.

Question

Accepted Answer

everyone in this video. I'm gonna go ahead and calculate for the ph and osmolarity of all the species from H three A. S. 04 All the way to O H minus were given the density and the equilibrium constants from K one K two and K three. Let's first find the clarity of H three A. S. 04 solution Using its density and mass percent. We go ahead and assume that we have a one liter solution. So for the mass of solution again we have one leader of this. Gonna go ahead and quickly convert this into milliliters. So for every 1000 mL we have one leader and then we're gonna go ahead and multiply this by the density. So we have 1.1033 g for every one millimeter. So for unit cancelation here We can see that the leaders will cancel and millions will cancel leaving us with the mass of the solution. To be one point or 1103.3g. The cackling for the mass of H three A S. 04. And solution, We're going to first start off with the mass of the solution. So that's 11:03.3 g of solution. We go and multiply this by the mass percent of the solution. That's 9.50 g per every 100 g of solution. We see then that the units of grams solution will cancel leaving us with the mass of the H three A S 04 in solution to be 104.8135 g. Alright. Now coupling for the moller mess, we're finding the molder mass rather of H three A. S. 04. Of course. Using the atomic masses from pr table or just finding this out on the internet. This is equal to one for 1. g per mole. Finally solving for the concentration or the more clarity of the solution is equal to 104.8135 g of H three A. S. 04 for everyone. Leader of solution we go hand multiply this by the molar mass. So for every one mole of H three A. S. 04, this gives us 141. g. So we see here for unit cancelation that the grams will canceled leaving us with just the moles over liters which is the exact units for more clarity. So once I put all these values into the calculator I get that the polarity of my solution is equal to 0. 375 Molars. Now for my first animation equilibrium scroll down for more space here. Alright 2nd. We're doing our first equilibrium, we have the reaction to B. H three A. S four Plus H two L. Which is of course a liquid. And this yields H30 plus and H two A. S. 04 minus. So we're basically gonna do an ice table here. So we have I for initial C. For changing the equilibrium. Of course these all concerned concentration for initial concentration of our first product here, our first set of material here, 0.738375 molars. For HBO this is liquid. So we'll go ahead and neglect all these values For our first product, this is zero as second product is also zero. So now for my change this is all unknown soldier known as X. From my study materials, there will always be an attraction for my products. This will always be a plus so addition. So with that being said for my first one here, that's minus X. For my two products, that's plus X. And plus X. For each of my products. Now for my equilibrium here basically combining these both. So for my H. Three A. S. +04, this is just 0.738375 molars minus X. Or two products. These are both excess. Now for our K. One value that's solved to be H. 30. Plus or rather the concentration of H. 30. Plus multiplied by the concentration of H. Two A. S. 04 minus divided by the concentration of H. Three A. S. 04. So we know some of these values of the school head plug that in. Starting with the K. A. One value that's given to us to be 6.46 times 10 to the negative three. Then we have the two vice here the numerator to be X squared because it's just X. Times X informative nominator. That is 0. molars minus X. So here I'm gonna go ahead and attempt to solve for X. So further I wanna go ahead and isolate my X squared so that we get 6.46 times 10 to the negative three times 0.738375 minus X. Equaling two X squared. We'll see that if we subtract X squared over we get a quadratic equation so that's X squared plus 6.46 times 10 to the negative three X minus 4.7699 times to the negative three equaling to zero. So we can go ahead and use the quadratic formula. Then we can get that My value of X is equal to 0.6591. As well as the X. Value to be equal to zero or negative 0.7237. Of course of course we know that the concentration cannot be negative. So this value of X will be canceled out meaning that concentration of H. Three A. S. 04 is equal to 0.738375 minus X. Which is just now figured out to be 0.6591. We've put that into a calculator received that. My concentration is equal to 0.672 molars now again scrolling down for more space, the concentration of H 30 plus is also equal to the concentration of H two S. 04 minus which is just equal to X. And we should solve that. And that's going to be 0.659 moller. So let's go ahead and now move on to our second equilibrium. Of course I'll do this in a different color here. So now we're dealing with the second equilibrium. So the equation for this here is h. two A. S. 04 minus reacting with H two A. H 20. Which is a liquid. Of course this is an equilibrium with the two products of H. 30. Plus. As well as H. A. S. 042 minus again do ai stable here. I for initial C. For changing the for equilibrium. So for this initial concentration we have zero point 0.06591. Of course water will all be neglected here. H. 30 plus is 0.6591 and zero. Now of course changes. Also still unknown. So we'll do notice this as y. For our second equilibrium. So if minus Y plus Y. And plus Y. Now for my equilibrium we're gonna go ahead and combine the initial and the change so equilibrium for my first time material is 0.6591 minus Y. For my H30 plus, this is 0.06591 plus why. And then for my second product that's just simply why. Alright so for mike a. To value is equal to the concentration of H. 30. Plus, multiplied by the concentration of H. A. S. 042 minus, divided by the concentration of H. Two A. S. 04 minus. Alright so we know the value of K. Tube since given to us in the problem. So let's just plug in all the values that we do know from the ice table as well as what's given to us in the problem. So that's 1.14 times 10 to the negative seven equaling to 0.6591 plus Y. Multiply it by Y over 0.6591 minus Y. So since the value of K. T. Is very small, we can assume that Y will be negligible. So you basically your head to cancel out this Y. Here and this Y. Here which leaves us 1.14 times 10 to the negative seven. Equaling to 0.6591 times Y. Divided by 0.6591. So now it will be much easier to solve and isolate for Y. Which we get the value then of 1.14 times 10 to the negative seven. This gives us the concentration then of H. A. S. 042 minus that equals to 1.14 times 10 to the negative seven molars. So now we can go ahead and move on to the third equilibrium. So scrolling down here again, we'll do this in a different color. So now we're dealing with our third equilibrium so same exact process here. The equation that's going on is that we have a church A. S. 042 minus reacting with H 20. Which is a liquid. This yields H. 30. Plus and a. S. 043 minus. Of course we have the I. C. And E. For initial concentration we have 1.14 times 10 to the negative seven. H. Show will all be neglected. H. 30 plus is 0.6591. And the last and final product is just zero. Of course the change is still unknown. So we'll denote it this time. For the third equilibrium to be easy. So I have minus C. Plus C. And plus Z. So from an equilibrium that's 1.14 times 10 to the negative seven minus C. For H. +30. Plus this is 0.6591 plus C. And my last product here is just Z. So now for my K. A. Three value that is equal to the concentration of H. 30. Plus, multiplied by the concentration of A. S. 043 minus divided by the concentration of H A. S. 042 minus. Alright again, we'll go ahead and plug in the values that we do have and know. So this then 3.16 times 10 to the negative equaling to 0.6591 plus Z multiplied by Z, divided by 1.14 times 10 to the negative seven minus Z. So since the value Of K. three is very small, we can assume that Y will be negligible. So basically we're gonna go Hamilton or cancel the sea and the sea here. So then we get 3.16 times 10 to the negative 12 equal to 0.6591, multiplied by Z divided by 1.14 times 10 to the negative seven. Go ahead. Isolate for a Z. Which is equal to 5.47 times to the 9 18 molars. Which then we already said is equal to the concentration of H A. S. 043 minus. And that is or rather our products, I apologize as the concentration of a. S. 043 minus that is 5. times 10 to negative 18 molars. So then we can go ahead and use this to find the ph value. So I get to my ph value is equal to the negative log of the concentration of H 30. Plus. We can just go ahead and plug in our value of Z. Which we can go ahead and plug in for the equilibrium. Or the concentration of H30. Plus. So that's going to be equal to the negative log of 0.06591. Once you put that into a calculator received at the ph value is 1.181. Now we can go ahead and find the concentration of H. Well scroll down for more space. So getting me to find concentration of O. H minus. So we have the concentration to be 0.6591, multiplied by the concentration of O. H minus. Eagling two Kw. So that is 1.0 times 10 to the negative 14. So this value earlier here, that's just the concentration of our H. 30 plus. Alright, so we're isolating our concentration of O. H minus between divide both sides by this concentration here, this gives us that the concentration of O. H minus is equal to 1.52 times 10 to the negative 13 molars. So they're running out all of our final answers for this problem, We see here that the concentration of H. three, A. S. 04 is equal to 0.672 molars. The concentration of H two, A. S. 04 minus is equal to 0.0659 Molars. My concentration of each A. S. 042 minus is equal to 1. times 10 to the negative seven molars, my concentration of a. S. 043 minus is equal to 5.47 times 10 to the negative 18 molars, my concentration of H 30 plus is equal to 0.659 molars. My concentration of O. H minus is equal to 1.52 times 10 to the negative 13 molars, and lastly, my value of P h is equal to 1.181. So all of these here are going to be my final answers for this problem.

A 7.0 mass % solution of H3PO4 in water has a density of 1.0353 g/mL. Calculate the pH and the molar concentrations of all species present (H3PO4, H2PO4-, PO43-, H3O+ , and OH-) in the solution. Values of equilibrium constants are listed in Appendix C.

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Key Concepts

Concentration and Molarity

Acid-Base Equilibria

pH and pOH Calculations