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(6) Calculate the volume of buffer to add t0 a lactase tablet that is 9000 FCC units to make solution al 1Q00 unitsImL concentration....

Question

(6) Calculate the volume of buffer to add t0 a lactase tablet that is 9000 FCC units to make solution al 1Q00 unitsImL concentration.

(6) Calculate the volume of buffer to add t0 a lactase tablet that is 9000 FCC units to make solution al 1Q00 unitsImL concentration.



Answers

What volume of 0.200 M HCl must be added to $500.0 \mathrm{mL}$ of $0.250 \mathrm{M} \mathrm{NH}_{3}$ to have a buffer with a pH of $9.00 ?$

Okay, so let's start by figure out how many moles of each substance we have here. Okay, so we have to moller And .355 L. So that's going to give us .0710 malls of the weak acid which is bicarbonate and 134 more and the same volume. Which will give us .04 76 moles of its conjugate base carbonate. Okay, so we have a weak acid and its conjugate base. So that is going to be a buffer. So our relationship for buffers is H plus is okay times the malls of acid provided by the malls of base. So that will be 47. I'm standing negative 11 some walls of acid over moles of base. So this will give us 7.0 I'm sounding negative 11 moller that's our H plus concentration. So if I take minus the log of that, we'll get the ph And that will be 10.15. In the second example we're adding some strong acid. That strong acid is going to react with the week based part of the buffer to make some more weak acid. So let's just do a table where we keep track of the strike geometry, initial change and final. So .03 Is what we're adding 2.04 76. and don't forget you have .071 of this to start the age plus is are limiting reactant. So we'll go ahead and subtract And add .03. She'll give a zero of that. .0176 moles and .101 moles. So we have a weak base and its conjugate weak acid. So that's a buffer. So let's go ahead and use our buffer equation H plus is a K a 47 times 10 to the negative 11 Acid over base. Which is our .101 Over .0176. So that's going to give us 2.7 Times 10 to the negative 10 more age plus. So We take minus or log of that. We'll get our ph which is nine 0.57. And then in this last example we're gonna add strong base. So the O H minus is going to react with the weak acid part of our buffer to make some water and some carbonate. So again we'll just keep track of this sort of in a store geometry table. So we're adding 0300 .0710. Don't care about the water. .0476. Okay. R O H minuses are limiting reactant so we'll subtract here and add over here So that will give us zero and zero for 10 And 0776. So this again is going to give us a week acid and its conjugate weak base so we have a buffer. So we'll do HPE plus is a K. A. Times acid over the base, Which is our .0410 over .07 76. So this gives us an h plus concentration of 25. I'm saying the -11 moller H plus. And then if we take minus or log of that, you've got your ph your ph becomes 10.60.

To determine the volume of one Moeller each sale that needs to be added to 750 milliliters of 7500.5 Moeller h peel four to minus. In order to reach a pH of seven, we can use the Henderson hassle Bolt equation. Our target pH is seven. That's going to be equal to PK or the negative log of the K A. That's going to be govern the buffer solution, which will be if we have HP will for two minus and we are adding HCL. We're going to create hte to peel for minus. Thus we'll have a buffer solution governed by PKK too. So we'll have K two here, plus the log of the moles of the base, the moles of the base. We start with the base being H. P +04 to minus is going to be 750 milliliters or 7500.75 zero leaders multiplied by the concentration. That's the molds we start with. But then when we add our unknown volume of HCL X at a polarity of 1.0 it'll decrease the moles of base by that same amount, and then we will create our moles of asset to the H to peel for minus, which will be also be equal to the moles of the strong acid added. The rest, then, is algebra. We'll subtract 7.21 from both sides and then take the anti log of both sides. Then we'll multiply both sides by X, add extra ble sides and then divide by 1.617 and we get X equal, 2.233 leaders or 233 milliliters.

This question is moderately challenging. Although very short in its description, a chemist needs a buffer with a ph of 3.50. So how many milliliters of pure formic acid? A density 1.2 to 0 g for mill leader must be added two, ml of .0857 Molar sodium hydroxide. Well, because we need are preparing a buffer solution, we can use the Henderson Hasselbach equation. Let me show you how I set it up. We have a desired ph of 3.50. That will be set equal to the Peca of the acid that will comprise the buffer formic acid plus the log of the moles Of the Form eight that we need In order to have a ph of 3.50. all of the Form eight is going to be coming from the sodium hydroxide That once reacted with the formic acid we add will create the Form eight. So the molds of form eight will be the moles of sodium hydroxide which will be its volume. 325 million litres converted to Leaders 3 to 5 multiplied by its concentration. This is the malls of format that will be present in the buffer, All of it coming from the moles of sodium hydroxide. We then divide by the additional volume. So we need to add enough to convert all of the sodium hydroxide into Form eight and then we need to add some additional formic acid in order to get moles of excess formic acid that will give us this ph because remember a buffer solution has both the form eight and the formic acid. So we'll take that additional volume multiplied by the density To convert the leaders into g. The density is 1.2- zero g for mill leader or 1220 g per leader. Then once we have the grams will divide by the molar mass of formic acid to get molds formic acid. Now we can solve for X. Which is just the additional volume of formic acid. After we've added enough to convert the sodium hydroxide into for mate. So we do our little bit of algebra. Subtract this from both sides. Take the anti log base 10 isolate our X. And we get 1.95 times 10 to the -3 leaders of pure formic acid. That's the additional amount though. So the total amount is going to be the additional amount plus the amount required to make this many moles of for mate. Well let's calculate again that many moles of form eight and figure out what volume of formic acid is needed to make that many moles of form eight from the sodium hydroxide. The moles can be converted into grams by multiplying by the molar mass of formic acid. So again this is moles, multiplied by Mueller mask it grams and then use density to convert the grams of pure formic acid into leaders summing these two up Gives us 3.1 times 10 to the -3 leaders or 3.01 mil leaders of pure formic acid.


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