# Chemistry

Calculate the change in PH when 6 ml of 0.1 M HCl is added to 100 ml of a buffer solution that is 0.1 M in NH3 and 0.1 M in NH4Cl. And calculate the change in PH when 6 ml of 0.1 M NaOH is added to the original buffer solution.

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1. 100 mL x 0.1M NH3 = 10 millimols NH3.
100 mL x 0.1M NH4Cl = 10 mmols NH4Cl.
6 mL x 0.1M HCl = 0.6 mmols HCl.
6 mL x 0.1M NaOH = 0.6 mmols NaOH.

...........NH3 + HCl --> NH4^+ + Cl^-
initial...10......0......0.1
change..-0.6...-0.6.......+0.6
equil....9.4......0......0.7....

Use the equilibrium values from the ICE chart, substitute them into the Henderson-Hasselbalch equation and solve for pH.

For the addition of NaOH, it's the same concept BUT you want this equilibrium:
...........NH4^+ + OH^- ==>NH3
Post your work if you get stuck.

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2. I did it in moles but still the homework assignment website is not accepting these answers.
My work:
Initial PH = 9.24.
moles of NH3= 0.01
moles of NH4= 0.01
moles of NH3 left= 0.0094
moles of NH4= 0.0106
PH= 9.19

moles of Naoh= 0.0006
moles of NH4= 0.0094
moles of NH3= 0.0106
PH= 9.30

used HH equation in all, considering PKa of NH3= 9.25

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3. I suspect the problem is that the question asks for CHANGE in pH and you're giving the data base the pH. I think something like 1.128 which I would round to 1.13 is the answer for the HCl part. I didn't work the NaOH part.
pH = pKa + log(9.24/0.7).
pH = pKa + 1.128. I didn't calculate the final pH since the change is what you want.

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5. You're right. That isn't th right answer. You should have checked my work. The error I made is I wrote 0.1 mol for NH4^+ and it isn't that to start.
It's 10............10
....-0.6............+0.6
final.9.4...........10.6

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6. I got initial PH= 9.24
so delta PH after HCl= 0.05
delta PH after NaOH= - 0.06
could it be that I'm calculating the total volume wrong? should be 0.01+0.006 L?

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7. I just took a quick look and I'm glad I did. I think you are working it right. You see my last response gives a value of 0.05218 (it should be -0.05218); you may not have keyed in enough significant figures. Round -0.05218 to the right number of s.f. from your original numbers.
The other suggestion I have is that one of the answers may be right but the other wrong. Are you keying in both answers at the same time. That will tell you if that could be the problem. No, I don't think you can calculate the wrong volume; even if you put in a wrong volume it won't make any difference if you use the same volume for both acid and base. That's why I work in millimols because mmols/mL = M but since mL is the same for both acid and base the volume always cancels.
I worked the base addition and the answer is the same except for a + number.
...........NH4^+ + OH^- ==> NH3
initial.....10...............10
change......-0.6..............+0.6
equil......9.4...............10.6
log b/a = log (10.6/9.4) = 0.05218
I have omitted the subtraction step since the initial pH is
pH = pKa + log(b/a) = pKa + 0 so the difference between pH of one soln and the original is just the log b/a term and I need not go through the subtraction step. But note this and this could be the problem. One of those 6 mL additions causes a +0.05218 pH difference and the other 6mL addition causes a -0.05218 pH difference. Are you keying in the negative sign? If you are allowed only 1 s.f. (from the 6 mL but I don't know if you just didn't type in all of the zeros) then +0.05 and -0.05 would be the answers. Parenthetically, I obtained 9.197 for the 9.19 you have and that rounds to 9.20 and would account for you obtaining 0.05 for one answer and 0.06 for the other. I'm going with 0.05 for one and -0.05 for the other if the 6 mL addition allows just 1 s.f. for the answer.
Please post a NEW post to me at the top of the page when you get it done. This page is getting so far back it's hard to find.

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8. You are right, thank you so so much.
With addition of HCl, the Ph= -0.052.
With addition of NAOH, the PH= 0.052.
Thank you for all your efforts.

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9. With pleasure.

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