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March 26, 2017

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1. A good indicator will have a(n)____ close to this volume and pH.
I think it's endpoint but it's wrong.

2. (part a) What is the pH of the solution created by combining 1.70 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?

2 (part b)What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water?
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my work so far:
(1.7e-3 L)(0.10 M)= 1.7e-4 mol NaOH
(8e-3 L)(0.10 M)=8e-4 mol HCl
(8e-4 mol Hcl)-(1.7e -4 mol NaOH)=6.3e-4

I don't know what to do next.

  • Chemistry - ,

    I did this last night for another and for different numbers. Let me try to find that and post it here. That will keep me from working it again. Actually you have two problems; i.e., one with NaOH and HCl and the other with acetic acid and NaOH. The second one is a buffer problem. I'll look for the other post I did last night.

  • Chemistry - ,

    http://www.jiskha.com/display.cgi?id=1335575263

    Let me know if you still have problems.

  • Chemistry - ,

    total volume=9.7e-4L
    .........HCl + NaOH ==> H2O + NaCl
    initial.8e-4....0........0......0
    add...........1.7e-4................
    change -1.7e-4 -1.7e-4 +1.7e-4 +1.7e-4
    equil..6.3e-4...o......1.7e-4..1.7e-4

    HCl:(6.3e-4 mol)/(9.7e-4 L)=0.6495 M
    and for pH I got 0.187 but it's wrong.
    I got the second part with acetic acid right though

  • Chemistry - ,

    also for 2. (part b) I found through M1V1=M2V2 that M2=8e-3 M acid. Then I did this: (8e-3 L)(8e-3M)=6.4e-5 mol acid. Next, I did the ICE table again and subtracted 1.7e-4 from 6.4e-5, and got -1.06e-4. I think I'm doing this wrong.

  • Chemistry - ,

    Sine you have part 2 I'll not go with that but here is part 1.
    I don't like mols for these small numbers because of all of those zeros that go with it. I THINK you 0.6495 should be 0.06495 but here is how to do it.
    1.70 mL x 0.1 M = 0.17 millimols NaOH
    8.00 mL x 0.1 M = 0.8 mmolsl HCl

    0.8 mmHCl - 0.17 mmols NaOH = 0.63 mmols HCl in excess (all of the NaOH is consumed). volume 1.7 mL + 8.00 mL = 9.70 mL.
    So (HCl) = 0.63 mmol/9.7 mL = 0.06495 M for which the pH is 1.187 which should be rounded to fit the problem. One problem with using moles with small numbers like this is that it's easy to lose or pick up an extra zero along the route.

  • Chemistry - ,

    Thanks, I understand the first part now but I keep getting the dilution part wrong.

  • Chemistry - ,

    Frankly, I don't understand the dilution part. I'm assuming that it means you take 8.00 mL of the 0.1M acid, dilute that to 100 mL, then add ALL OF IT to the base. If that is the right meaning, here is how it's done.
    1.70 mL x 0.1 M NaOH = 0.17 millimoles.
    8.00 mL x 0.1 M HCl = 0.80 mmols HCl
    total volume = 1.70 + 100 = 101.7 mL.

    Now (HCl) = 0.63 mmols/101.7 mL = 0.00619 and pH = 2.2 which is about a tenth of the first value.

    For the acetic acid.
    1.70 x 0.1 M NaOH = 0.17 mmols.
    8.00 x 0.1 M HAc = 0.8 mmols HAc
    total volume = 1.7 + 100 = 101.7
    (HAc) = 0.63/101.7 = 0.00619 M
    (NaAc) = 0.17/101.7 = 0.00167 M
    pH = pKa + log (NaAc)/(HAc)
    pH = 4.74 + log (0.00167/0.00619)
    pH = 4.74 - 0.57 = 4.17.
    If this is the right way to interpret the problem, then note that the addition to the strong acid/strong base mixture (not a buffered solution) caused a change in H^+ by a factor of 10 (0.06 to 0.006 or pH about 1.2 to 2.2) BUT the added water to the acetic acid/NaAc final mixture (a buffered solution) didn't changed the pH at all (from 4.17 to 4.17 if you used the same Ka I used). I think the lesson here is to see the difference in how dilution with about 92 mL water didn't change the pH at all for a buffered solution but changed significantly for an un-buffered solution).

  • Chemistry - ,

    I see what I was doing wrong now. Thank you very much!

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