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Iwant to makc 1.0 litcr of a buffer that will kccp the pH of a solution at 7.40. If I start by dissolving 30 g of sodium dihydrogen phosphate in watcr; how many mil...

Question

Iwant to makc 1.0 litcr of a buffer that will kccp the pH of a solution at 7.40. If I start by dissolving 30 g of sodium dihydrogen phosphate in watcr; how many milliliters of cither 0.50 M HCL, or 0.SOM NaOH will have to add to this solution to makc my buffer? (6 pts) For phosphoric acid: Kap 7.11 *10" Kaz - 6.32 * 10 ; Kaz - 4.5 * 10-13

Iwant to makc 1.0 litcr of a buffer that will kccp the pH of a solution at 7.40. If I start by dissolving 30 g of sodium dihydrogen phosphate in watcr; how many milliliters of cither 0.50 M HCL, or 0.SOM NaOH will have to add to this solution to makc my buffer? (6 pts) For phosphoric acid: Kap 7.11 *10" Kaz - 6.32 * 10 ; Kaz - 4.5 * 10-13



Answers

You have to prepare a $\mathrm{pH} 4.80$ buffer, and you have the following $0.10 \mathrm{M}$ solutions available: formic acid, sodium formate, propionic acid, sodium propionate, phosphoric acid, and sodium dihydrogen phosphate. Which solutions would you use? How many milliliters of each solution would you use to make approximately a liter of the buffer?

So in this problem we have this n a h c 03 and this any to ceo three. And it gives us how many grams we have each and we're trying to find the pH. And so first we need to get these into moles, and then they're hilarity. So 4.2 grams off this n a h e 03 and we can look up. You look it up or using the PR table, we find a smaller master B a four point 007 grams per one mole. And that equals 0.50 moles after some rounding. And then you divide that by how many it leaders it tells us which we have, which is 0.20 leaders. And that equals 0.25 Moeller now do the same thing for the other one for the other, and a to C 03 We have 5.3 grams and to convert it with these molar mass again and we can look that up or use your table to solve. And that's 105.98 grams per one mole. And that gives us again approximately 10.0 point 050 moles and we can divide that by the same leaders could there in the same solution. And we get the same mill Arat e. And now we can, you know, as the hundreds of household Bach equation to find the pH. And as you guys already can see, the log term will be one. But all right, it anyways, So the P ph this is demonstrating this process. Bach is the PKK, plus the log of the base over the acid and with the find is a PK value. And luckily, you could either look up PK a values or use the back of the book. But I use the back of the book here, and you just take the negative log off the PKK to, because this is the second generation off the acid of the because it's of acid because it can be HTC 03 for the carbonic acid and negative log of PK two is equal to 10.25 10.25 and therefore we get Ph equals 10.25 plus the log of this 0.25 Moeller divided by a 0.25 Miller. This just goes to zero because the log of 10 So you just get the pH equals P care equals 10.25 And now it's asking us, does the pH change if you double the volume of the entire If you change the volume times 10 and the answer is no, it still stays the same because if you divided both of these 0.5 bowls by the by any of the same numbers, if you put 100 here 300 here, if you do it to the bold operations, you get the same number for each. So that means this this ratio will always equal one. So you just get a pH of 10.25 At a certain point, however, if you add too much, you will too much water. The water of the pH will start to dominate, therefore, may be making the pH differ. But say if you increase by a factor of 10 pH will stay the same with this 10 25

Solution for the above problem here. Lover pH means in Greece. Acidity. Saw your in Greece. Concentration off concentration off at C two is three or two are decrease concentration off C two h five or too negative here on Lee, it's hell will rise. Hydrochloric acid will rise. Concentration edge C two is three or two. Notarized the concentration off H C two H three or two and pH physical to five physical to 4.74 plus log into concentration. Off. C two is three or two negative, divided by concentration off edge. See too. His three or two is equal to 1.8 and you number of malls off C two at three or two. Negative is equal to 300. Emmel, divided by 1000 multiply 0.560 is equal to 0.168 moles and number of yourself. C two is three or two. Etch is equal to 300 divided by 1000. Multiply 0.25 is equal to 0.75 moles and 1.8 is equal to 1 68. Minus eggs divided by divided by year 15 plus eggs. Eggs is equal to 12. Mall off it three or positive and volume off 0.15 Malone Ity It's sealed is equal to wealth. Multiply one upon 0.15 is equal to 80 on how.

For this problem. It is rather complicated, although simple in its description, let me see if I can walk you through it. So we want to have a buffer solution with a ph of 4.35 by adding pure acetic acid to 465 mil. Leaders of a .0941 molar sodium hydroxide solution. Once the buffer solution is created we will have a certain number of moles of the acetate created from the sodium hydroxide. When we add acetic acid to the sodium hydroxide plus some additional pure acetic acid after being added to the sodium hydroxide. That is more than the sodium hydroxide. So that when we're all done we have some acetate created from the sodium hydroxide reacting with acetic acid and some additional acetic acid. Thus we have a buffer solution of acetic acid and acetate. So we can use the Henderson Hasselbach equation. Ph the desired ph we want will be equal to the Peca of the acid. In the buffer the acid is acetic acid. So we look up the ca of acetic acid and take the negative log of that value plus the log of the moles of acetate. The moles of acetate that we will have in the buffer solution will come exclusively from the sodium hydroxide. Whether acetic acid. We add reacts with the sodium hydroxide, creating acetate. So the moles acetate will be the volume of sodium hydroxide, 465 ml or 65465 Leaders multiplied by the concentration of .0941 volume and leaders times more clarity Always gives us moles divided by the moles of formic acid. The Mosul formic acid will be created after converting all the sodium hydroxide into acetate. So this will be the additional volume. X. Is the volume of acetic acid added. After all of the sodium hydroxide has been converted into acetate. That volume. When multiplied by the density of acetic acid will give us grams acetic acid which we can then divide by the molar mass of acetic acid to get molds acetic acid. So now we have a single equation for which we can solve for the moles of acetic acid. I'm sorry the volume of citric acid that needs to be added after adding enough to convert all of the sodium hydroxide into acetate. We'll subtract this value from both sides. Take the anti log base 10 from both sides and then isolate our X. Value. We get 658 times 10 to the negative. Three leaders. Now remember this is the additional amount we need to add that to the amount required to convert all of the sodium hydroxide into acetate to do that will calculate the moles of sodium hydroxide as we did up here by taking the volume and leaders multiplied by be more clarity when we have mold sodium hydroxide we recognize that that will be equal to the moles of acetate formed the most of acetate can be converted to graham's acetate by multiplying by its smaller mass and then convert the grams acetate into leaders acetate by multiplying by, I'm sorry, by dividing by the density expressed in grams for leader rather than grams per mil leader, which was provided in the question. The sum of these two values gives us the total volume of acetic acid that is needed 9.09 times 10 to the -3 L or 9.09 ml. That's pure acetic acid.

First let us Cal Khalid B K A. For this value. So the foreigner we are using its peak a is equal to minus off. Log off. Hey, so the final be care value is seven point for six. Ex used the Henderson aggression to solve the it This is the hand disintegration immigration. After substituting the values in the creation, we get a B it off eight point cool. Then he h is equal to the value off B A. What offer? Transit fiction off acid is equal to the concentration off this here. The acid concentration needs to be increased by the base. Concentration needs to be decreased in order to get both have equal concentration. What's your problem? More asset is needed to be added in the buffer solution for the last, but even in the table, our initial concentration Auntie changing concentration after adding the acidic US and at P h equals toe peek A. Both asset and bass have same concentration. Hence we can solve that and get value off X, which equals Toby. Zero point 35 m. Therefore, 0.35 mole off asset. It still must be added to initial buffer. The concentration or a said the concentration off acid on base eventually equal to one another.


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