Neutralization reactions involving either a strong acid or a
strong base go essentially to completion, and therefore we
must take such neutralizations into account before calculating
concentrations in mixtures of acids and bases. Consider
a mixture of 3.28 g of Na3PO4 and 300.0 mL of 0.180 M
HCl. Write balanced net ionic equations for the neutralization
reactions and calculate the pH of the solution.

Question

Neutralization reactions involving either a strong acid or a 
strong base go essentially to completion, and therefore we 
must take such neutralizations into account before calculating 
concentrations in mixtures of acids and bases. Consider 
a mixture of 3.28 g of Na3PO4 and 300.0 mL of 0.180 M 
HCl. Write balanced net ionic equations for the neutralization 
reactions and calculate the pH of the solution.

Accepted Answer

Welcome back everyone. We're told that neutralization reactions involving a strong acid like hydrochloric acid proceeds to completion. A four point oh five g sample of tri sodium citrate was dissolved in 275 mL of 2750.1 60 molar hydrochloric acid. Right the bounce net ionic equations for the neutralization reactions involved find the ph of the resulting solution and were given the acid dissociation constant K. A. For citric acid as 7.41 times 10 to the negative fourth power. Recall the formula for citric acid being C6 H and focusing on our weak base, which we recognize as try sodium citrate, recall that a weak base does not fully dissociate. And so when we dissolve our tri sodium citrate salt in water, we would form our situate. Try an ion with a three minus charge and we would form three moles of our sodium Catalans which are Spectators. Now our citrate, try an ion is our weak base here, which will not fully dissociate. And so it will be undergoing in equilibrium in the neutralization. Now we need to recognize that the prompt tells us our mass of tri sodium citrate as 4.05 g and we need to find the moles. So we're going to multiply by the molar mass of tri sodium citrate from our periodic table And calculating using the chemical formula and the masses in our periodic table for each of the atoms in this formula, we would find a mass equal to 258.069 g of tri sodium citrate equivalent to one mole of tri sodium citrate, canceling out our grams of tri sodium citrate. We are left with moles and we would come up with a value of 0.15693477 moles of tri sodium citrate. But we need to or we can actually make this an even smaller unit by converting to millennials. So we would multiply by the conversion factor to go from moles to millennials where our prefix milli tells us that for one mole we have an equivalent of 10 to the third power millennials. So canceling out moles of tri sodium citrate, we would calculate, I'm sorry this says mel mel's here. We would calculate a total of 15.6934 mila mols of tri sodium citrate. And because we see that we have a 1-1 molar ratio between tri sodium citrate and are situated. Try an ion. We can say that therefore our mila mols of tri sodium citrate are equivalent to our mila mel's of our weak base. So equivalent to our millennials of situate, try an ion. Now let's focus on hydrochloric acid, which we would recall is actually a strong acid. And so it therefore should fully dissociate. And so we would form when we dissolve hydrochloric acid and water. We would form the chloride an ion and we would form hydro ni um recognize that hydrochloric acid acts as a bronze ted. Lowry acid donating a proton to water. Water being our bronze did larry base accepts this proton and we form again our products, chloride and hydro ni. Um Now from the prompt we are told that we have a Volume of hydrochloric acid being 275 ml And were given the concentration in as .160 Moller. And so here we would multiply by recalling that we can interpret our polarity in terms of mila moles per millimeter. And so we would plug in that we have 0.1 60 millimeters of hcl equivalent to one mL of solution and simplifying this into our calculators, we would cancel out milliliters of hydrochloric acid were left with mila moles of hydrochloric acid And we would yield a result of 40 for Mila Mel's of Hcl now recognize that chloride is the conjugate base of our strong acid, hydrochloric acid and hydro ni um is going to be the conjugate acid of our base being water. So because of that, we want to look at the ratio between our strong acid and our conjugate acid. So in this case we're looking at the ratio between hcl and hydro ni. Um and we would say that we have coefficients of one. So we have a 1 to 1 ratio and so therefore the mila moles of hydrochloric acid are equivalent to the millennials of hydro ni um being both 44. Now based on what we've determined so far, we can see that we have only 15.69, 77 Mila Mel's of our weak base. Try acetate an ion. And we have 44 millennials of our conjugate acid, hydro ni. um So we would say that our limiting reactant is going to be our tri acetate. Sorry, are try citrate an ion. So now let's go ahead and write out our first neutralization ionic equation. So we would begin with our acid hydro ni um which is going to be neutralizing our weak base. Are try citrate an ion. Or sorry situate, try an ion. So C six H -. This is an equilibrium because again our citrate trianon is a weak base and does not fully dissociate now because hydro knee um is our acid. We would have it acting as a bronze stead acid. So it's donating a proton to our situate. Try an ion and we would as a result form C six H 607 to minus and water. So again hydro knee um is our bronze dead Lowry acid. Our citrate ion ion is our bronze ted Lowry base accepting that proton, our C six H 6072 minus an ion is our conjugate acid and water is going to be our conjugate base in this case as a product because we understand that our citrate. Try an ion is our limiting reagent. Because we have a lower amount of millennials of it initially. So we'll fill that out. We have initially 15.69 34 77 millimeters present initially, which we determined above. And initially for hydro ni um we have 44 millimeters present since citrate trianon is our limiting reactant. The change is going to be minus the amount of our citrate trianon so minus 15.69 30 for 77 for both of our reactant. And the change for our products, which would only include our acquis product would be plus 15.69 34 77 millennials. Now we can write out the equilibrium concentrations where for hydro ni um we would have 28. 23 millennials for citrate trianon and we would have none left because it's fully consumed. And for R. C six H 6072 minus an eye on we would have 15.69 34 77 millennials. And again, water is not considered it's a liquid. Now this first neutralization that we've written out is going to be our first expression here, specifically the first expression of our net ionic equation in the neutralization. Now this neutralization continues until we form citric acid and we noted that the formula for citric acid is C six H 807. So we need to continue until we get that as a product. So now we need to make a second ionic equation where we begin with C six H 6072 minus reacting with hydro ni um which are what we have left over at equilibrium. And this is an equilibrium where again hydro knee um acts as our bronze stead. Lowry acid donating a proton to our C. Six H 6072 minus which is our bronze ted. Lowry base, accepting the proton. And as a result we form the product where we have C six H seven 07 -1 as our first product. And then we would have water where again water is our conjugate base of hydro ni. Um and C six H minus one is our conjugate acid of our base. And so now let's go ahead and note down the initial concentrations we have for C six H 607 minus two. A initial concentration of 15.69 34 millennials for hydro knee. Um We have an initial concentration of 28. millennials. Our products are zero. Initially the change is going to be the full consumption of our base. And so we would have for the change minus 15.69 77 for both our reactant and for our products we would have plus that change we disregard water yet again. So sorry for including the zero there but we would have for the change of our conjugate acid plus 15.69 34 77. Now for our equilibrium concentrations. So this is an E. Here we would have the results being zero left of our base. We would have for our hydro ni um a remaining concentration of 12.613046. And for our conjugate acid product we would have again Plus 15.6934 77 at Equilibrium. Now we have written out our second net ionic equation for the dissociation. We're sorry for the neutralization of our tri sodium citrate. Now we're going to continue on because we still need to form our citric acid as a product. And so we would begin with yet again we have hydro ni um which initially is 12.6130 46 millennials present reacting with our base which is now in this case C six H 7071 minus which has an initial concentration at 15. 34 77. This is an equilibrium and we will form yet again because our hydro knee um access the bronze. Did larry acid. We donate a proton. And our basic steps this proton and we would form The conjugate acid being C6 H8 07 which is our citric acid. So that's good. We know we end here and then we have our product left over which is water as our second product. We disregard this in our equilibrium chart. Now to determine our final concentrations. Again, our change is going to be minus the lower amount. And so in this case now we have less hydro ni um than our weak base. And so we would have the change now being minus 12.613046 millennials for both of our reactant And for our product we have plus this amount. Now adding this together, we would have at equilibrium a concentration of zero for hydro ni um for our weak base we would have a difference of three point oh eight oh 431 at equilibrium. And then for our citric acid we would have an equilibrium concentration of 12.61 3046. Now recognize that our citric acid is our conjugate acid of our acid being hydro nia are bronze. Did larry acid specifically since it donates a proton. And so because our hydro ni um as we stated, has an equivalence in Milham als to the concentration of or mila moles of hydrochloric acid, we can say that our volume of hcl is also going to be equal to our volume of citric acid. And so therefore we can determine the polarity of citric acid by taking the equilibrium amount which we determined as 12 point sorry, we've determined that as 12.613046 Mila mel's of citric acid and we're going to multiply this by our we're sorry, we are converting to more clarity. So we are going to divide this by our volume of of hydrochloric acid which is A K. A. Our volume of citric acid. So we're dividing by the given volumes from the prompt for hydrochloric acid. As 275 mL. This is going to result in our polarity of citric acid equal to a value of 0. moller for citric acid. Now note that we also have left over at equilibrium some of our bronze dead Lowry base. Sorry, that's labeled up here which was our proton except er and so we need to calculate the concentration of that base. So For the polarity of C6 H707 we would take its equilibrium concentration three point oh eight oh 431 million miles Of C six H 707 -. And we are dividing by the concentration of our solution. Again being 275 mL. This results in a molar concentration equal to 0.112016 moller. So for our third answer being our third balanced net ionic equation here, we have our three net ionic equations highlighted in yellow but now we need to find the ph and now because we know our molar concentrations of our base and conjugate acid. We're going to show the the association of citric acid which we stated is a weak acid. So it's association or as it dissolves in water, it's going to be an equilibrium since it's a weak acid. So it's not going to fully associate. We must have an equilibrium arrow. And we're going to have our product B. R. Weak base, which is C. Seven or sorry, that's C six, Age 707 as well as water. We know that these are our products because we're sorry, our second product is hydro ni um We know that these are our products because in this case our citric acid acts as our bronze ted Lowry acid, donating a proton to water being our bronze Dead. Lowry base, accepting that proton, we form our conjugate base here and hydro knee. Um as our conjugate acid of water. Now recognize that citric acid has a total of eight protons in the formula. And recall that this is considered a multi product acid. And recall that a multi product acid will have a major effect on ph only from the first, only from the first dissociation. And so we're only going to Show this final ice chart where we have our initial concentration of citric acid which above is 0. Molar. For water, we would disregard it. And sorry, this should be a liquid symbol here Then for our conjugate base we have plus the we're sorry, we have 0.0112016 Mueller as our initial concentration and we don't know our concentration of hydro ni um initially. And so our change is going to be minus the concentration of hydro knee. Um which is minus X. In this case it would be plus X. For our conjugate base. So this is plus X here and plus X. For hydro ni. Um So carrying everything down for equilibrium, we would have 0. minus X. For citric acid. For our conjugate base. We would have 0.112016 plus X. And for hydro knee. Um We would just have plus X. Or rather X. In this case. Now to figure out the ph of this final solution, we need to find our concentration of hydro ni. Um So we need to solve for X. And so we're going to need to write out our K. Expression which we are told the K. A. For citric acid from the prompt as equal to 7.41 times 10 to the negative fourth power. And recall that K. Is equal to our concentration of products being hydro ni. Um And our conjugate base divided by our concentration of reactant. We're only focusing on citric assets since it's the only acquis reactant. So plugging in what we have from our chart above we would have 7. times 10 to the negative four of our K. A. Equal to our concentration of hydro ni. Um Which is X. Our concentration of our conjugate base 0.112016 plus X. At equilibrium divided by our concentration of acetic acid. A equilibrium 0.48656 minus X. At equilibrium we need to come up with a quadratic expression. So we're going to multiply by our denominator, 0.048656 -1 on both sides. 48656 -1. And this is an eight. And apologies. We want to make sure we have the proper value. So for going back to our step for the molar concentration, I misinterpreted my results and it should be 0.045866 molar. And so going to our ice chart, we would have citric acid as a concentration being 0.045866 Mueller. Initially for the equilibrium change we would have 0.45866 smaller minus X. And in our K expression, we would have That as well. So 0.045866 -1. And so we would be multiplying both sides by that. So apologies for that error but I'm glad I caught it. We have a long problem here. So let's keep going. And when we take the product of the left hand side, we would come up with the result being 3. times 10 to the negative fifth power minus 7.41 times 10 to the negative fourth power X equal to 0.112016 X. Multiplied by X. Now taking the product of our right hand side of our equation, we have 3.398641 times 10 to the negative fifth power minus 7.41 times 10 to the negative fourth power X equal to 0.112016 X squared. Or rather X. And then plus X squared. So this is a square power here. Next we want to combine our like terms, so our two X terms and we will come up with the final quadratic equation where we would get X squared plus 0.119426 X. When we take the some of our two like terms minus 3.398641 times 10 to the negative fifth power all equal to zero. And so A is going to be one B is going to be 0.119426 and C is going to be -3.398641. And so recall our formula for the quadratic equation where we have x equal to negative B plus or minus instead of our square root symbol. We can represent in brackets. The falling parentheses which would be B squared minus four times a times C. Raised to the 1/2 power which is all divided by two over a or two times a. Sorry. And so incorporating this using our terms, we would say that X is equal to negative 0.1194 to six plus or minus. We have our brackets are parentheses for b squared, so 0.119426 squared, subtracted from four times one times C which is negative 3.398641. And we need to include that times 10 to the negative fifth power, so times 10 to the negative fifth power. And just so this is visible. I'll scoot this over. So let's just make this X will rewrite that Simplifying This. We still have we have our brackets here, we're still dividing this entire quotient by two Which is multiplied by one. And so we would come up with in our calculators, 2X values. The first x value equaling negative zero point 0143165 for our first x value. And we would have a second X value calculated as positive 2.373928 times 10 to the negative third power. And of course we want to go with our positive X term. And so we can say that our concentration of hydro knee um is equal to X according to our chart which is equal to 2.373928 times 10 to the negative third power. And recall that we can calculate ph by taking the negative log of our concentration of hydro ni. Um and so plugging that in, we would have the negative log of 2.373928 times 10 to the negative third power Equal to 2.624532. Which we can round to exactly three sig figs as our minimum number of sig figs considering the info given in the prompt. So rounding this 23 sig figs, we would get 2.62 as our final ph of the solution. So what's highlighted in yellow for our ph of the solution as well as our balanced net ionic equations from above. We had three of those here highlighted our our final answers to complete this example corresponding to choice a in the multiple choice, I hope everything argued was clear. If you have any questions, leave them below and I'll see everyone in the next video

Verified Solution

Key Concepts

Neutralization Reactions

Net Ionic Equations

pH Calculation