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Point ) At 1(PC and [.00 atmn for the Tectoi H,O() = H,O(g). is AS positive. negative zero" points) We dissolve A H of A acid. HA elough water t0 produce 25.0 ...

Question

Point ) At 1(PC and [.00 atmn for the Tectoi H,O() = H,O(g). is AS positive. negative zero" points) We dissolve A H of A acid. HA elough water t0 produce 25.0 mL of solution_ Hle titrate this solution with 0.222 M solution of NaOH Audl rexquire 33.3 IL to Fench the eqquivalence pint Tle vlue of for HA(aq) is 8.8 (3 points) Wlmt is tle molar IIIARS of IIA?poiuts) Wlmtthe pH of tle HA(A4) solution bwfore we started tle titration? poiuts) Wlat WALY the p At the (Quivalence point?

point ) At 1(PC and [.00 atmn for the Tectoi H,O() = H,O(g). is AS positive. negative zero" points) We dissolve A H of A acid. HA elough water t0 produce 25.0 mL of solution_ Hle titrate this solution with 0.222 M solution of NaOH Audl rexquire 33.3 IL to Fench the eqquivalence pint Tle vlue of for HA(aq) is 8.8 (3 points) Wlmt is tle molar IIIARS of IIA? poiuts) Wlmt the pH of tle HA(A4) solution bwfore we started tle titration? poiuts) Wlat WALY the p At the (Quivalence point?



Answers

A $0.200-\mathrm{g}$ sample of a triprotic acid (molar mass $=165.0$ g/mol) is dissolved in a 50.00 -mL aqueous solution and titrated with $0.0500 \mathrm{M} \mathrm{NaOH}$. After $10.50 \mathrm{mL}$ of the base was added, the $\mathrm{pH}$ was observed to be $3.73 .$ The $\mathrm{pH}$ at the first stoichiometric point was 5.19 and at the second stoichiometric point was 8.00 a. Calculate the three $K_{\text {a }}$ values for the acid. b. Make a reasonable estimate of the pH after $59.0 \mathrm{mL}$ of $0.0500 M$ NaOH has been added. Explain your answer. c. Calculate the pH after 59.0 mL of $0.0500 M$ NaOH has been added.

Yeah. To solve for the volume of sodium hydroxide needed. We'll start with the volume of the acid, convert the volume two mil leaders. I'm sorry the volume in mill leaders to leaders by dividing by 1000. Then multiply by the polarity to get molds of the acid. We then recognize the reaction is 1-1. So we get mold sodium hydroxide and then we'll divide by the polarity of sodium hydroxide to get peter sodium hydroxide And then multiply by 1000 to get ml sodium hydroxide. So 59 ml of sodium hydroxide is required to carry out this type tradition. This is sodium hydroxide. Then to solve for the ph at the equivalence point the hydroxide concentration because a base has been created will be equal to the square root of the KB value which is going to be the K. A. Value for acetic acid divided into K. W. Multiplied by the concentration of the acetate that is present at the equivalence point which will be the moles of acetate which is the molds of acetic acid. We start with which is its volume multiplied by its polarity Divided by the new volume which is the 42 million L plus what we just added 59 mil leaders converted to leaders And we get 347 times 10 to the -6. This being the hydroxide concentration. We can then solve for P. H. By taking the negative log of the hydro knee um concentration which is the hydroxide concentration divided into K. W. And we get 8.54. Next for part B. We'll do the same thing. We'll start with the volume of the asset that we have convert from mill leaders to leaders than leaders. To mull with more clarity here, we recognize that we need two moles of sodium hydroxide for every mole of sulfurous acid. Well then convert the Mosul of sodium hydroxide to leaders by dividing by its polarity and then from leaders to mill leaders. So this is what's required to reach the second equivalence point half of that is what's required to reach the first equivalence point. The ph of the first equivalence point where we have the unflattering species. H S 03 minus can be calculated as ph equals one half PK one plus PK two. This is typically the best way to solve for ph when we have an ample Terek species And we get 452. Now to solve for the ph at the second equivalence point will have the soul fight and ion. And we can solve for the hydroxide concentration by taking the square root of The KB one value which will be the K two divided into KW. Multiplied by the concentration of sulfite at the equivalence point. The moles of H203 that we start with will now be equal to the moles of sulfite that we have which will be the initial volume multiplied by the initial polarity. We then divided by the new volume which will be the volume. We started with Plus 132 million liters, And we get a hydroxide concentration of 45 times 10 to the negative five ph will then be the negative log of the hydro knee um concentration, which will be the hydroxide concentration that we just calculated, divided into K. W. Giving us a ph of 969.

So starting with a here 18 02 plus in a oh, age will titrate. So age you know, two equals 25 times 0.132 which is equals a 3.3 more. And then in a oh, age is equal to 10 times your 100.1 when 16 which is equal to 1.16 mol and then h and no. Two is an excess. So use a Henderson HASA back equation and Ph Will Ego 3.14 plus log of 1.1 6/2 0.14 which is equal to 2.87 And then for B, we have aged, you know, two equals 3.3 wall in a reach is equal to 2.32 and then we have excess eight, you know, to logical 0.98 more. So then I use the same Anderson house about equation and this is our answer

The first thing that we should do in order to solve the various steps of this titrate Asian is writer overall equation, and for that we will have aged to a plus two moles of sodium hydroxide going to any to A plus two moles of water. And now the next step is we need to find the moles of our H two a. So you will take the volume of any wage, multiply that by arm polarity of anyway h and then multiply this value by our Miller ratio of one mole of H two a over two moles off sodium hydroxide. So what that will look like this. You will take our volume 34.27 And we know that one leader is 1000 with the leaders. And we have 0.1 moles of any of which, in my one leader solution. Multiply that by our molar ratio of 1/2 and we will get the molds that we need is 0.1736 And now we confined them overweight, which is this is denoting Mueller wait of h two a. You will take 0.2015 grams divide that by the moles we just calculated, and we will get 100 16.1 grams per mole. So a couple values that we need to know the volume of our asset is 0.25 liters. The volume of our sodium hydroxide is 34 point 72 milliliters. The polarity of of our sodium hydroxide is 0.1 Moeller. And to get the final volume, we simply add the volume of her acid and the volume of our base, which is anyway H. And that will be points here to five post 34.72 And we will get that. Our final volume is 0.572 leaders. Next thing that we need to do is set our pH equivalent to the average of our PK A values. So this means that at the second equivalents point, I have a different equation in my tie Trish in, I'll have a to minus plus water going to H K minus plus hydroxide, and we also know, based on the definition of our base ionization constant K B. And it's equivalent to K W, divided by K eight to which will be our each a concentration multiplied by her are hydroxide concentration divided by the concentration of eight to minus. In order to get the concentration of hydroxide, we'll take the square root of K W, divided by K two multiplied by the concentration of eight to minus. The next thing that we can do is find the concentrate concentration of our hija rhodium. And to do that we take 10. Raise the negative pH So we'll take 10 race negative. 9.37 and we will get that. Our concentration is 5.4 times 10 to the negative 10th and now we confined the concentration of our hydroxide. We'll take 1.0 time to tend race the negative fourth, divided by the concentration of the high Jeroen IAM. I will get 1.9 times 10 to the negative fifth and now we can take the final volume that we calculated previously. And we'll use that to find the concentration of a to minus. We'll take our the molds that we found before our initial divide that by our volume, and we will get that the concentration is point there to 907 Moeller And now we confined an equation for K eight to we'll take a one times 10 to the negative 14 times the concentration we just previously found points your 2907 divide that by 1.9 times 10 to the native Fifth Square. Again, this is a value we just previously calculated as well. And we will get 8.1 times 10 to the negative seventh. And now we confined the PKK that we need. We'll take a been negative log of 8.1 times 10 to the negative seven and we will get that our first. I'm sorry. Our second PK value a 6.9 and now defined our first peek A value will multiplier pH times two Subtract the second p k A, and we will get that the first PK a value is 1.81

Number 1 18 is simply a die protic acid titrate Asian calculation. The two pH is that air given really are not important to the calculation. We simply need to recognize that the sodium hydroxide has neutralized the dye protic acid with the addition of 31.50 mill leaders of the 0.984 Moeller sodium hydroxide solution. So because we recognized, then that the final failing indicator is going to change color close to Ph. Nine, the only purpose of the two pH values that they gave to us is recognizing that the second equivalence point is going to occur at Ph. 8.8 and therefore when final failing changes color were at the second equivalents. Point meaning sodium hydroxide has reacted with both the first and the second hydrogen ions on Milan IC acid. So we now know this, like geometry is 2 to 1. We need two moles of sodium hydroxide for every one mole of Milan ic acid to reach the second equivalents point so we can calculate the moles of sodium hydroxide, added convert two moles of melodic acid with the 2 to 1 relationship, then divide the most of Milan ic acid by the newt. The total volume which is going to be the original 25 mL of Milan ic acid before we added the sodium hydroxide. So this is not the some volume. This is just the volume of Milan ic acid before titrate Asian, and we get 250.62 Moeller melodic acid.


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