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March 26, 2017

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I posted this yesterday and tried to work it out today but i don't remember how to. You said convert to moles by multiplying for example .1M x .008L= 8x10^-4 ...after this what do i do with this number to find the pH i tried to just plug that into the -log(8x10^-4) but it was wrong.

What is the pH of the solution created by combining 2.30 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)?

Here is the chart that i am supposed fill out:

mL NaOH pH w/ HCl pH w/ HC2H3O2
2.30 _________ ____________

Here's the second part of the problem.
I need to complete the chart too

mL NaOH pH w/ HCl pH w/ HC2H3O2
2.30 _________ _____________

  • Chemistry -- to Dr.Bob - ,

    You have too many questions within questions which makes it confusing. Let me separate the first one.
    What is the pH of the solution created by combining 2.30 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)?

    NaOH + HCl ==> NaCl + HOH
    Strong base + strong acid.

    mols NaOH = M x L = 0.1 M x 0.00230 L = 0.000230 mols.
    mols HCl = 0.10 x 0.008 = 0.0008
    So 0.000230 mols NaOH react with the 0.0008 mols HCl to form 0.000230 mols NaCl and water, but 0.000800-0.000230 = 0.00057 mols HCl remain.
    What is its concn. That will be mols/L.
    We have 0.00057 mols and we have 2.30 mL + 8.00 mL = 10.3 mL or 0.0103 L
    So (HCl) = 0.00057/0.01030 = 0.05534 M.
    Since HCl is a strong acid, that is the (H^+) and pH = -log(H^+) = -log 0.05534 = 1.2569 which I would round to 1.26.


    For the NaOH + acetic acid here is what I said.
    Determine mols NaOH.
    Determine mols acetic acid.
    Determine how much of the salt (sodium acetate) is formed and how much of the acetic acid is left unreacted. (Same process s above).This mixture of acetic acid + sodium acetate is the salt of a strong base and a weak acid so that makes it a buffer solution. The pH of a buffer solution is determined by the Henderson-Hasselbalch equation.
    pH = pKa + log(base)/(acid)

  • Chemistry -- to Dr.Bob - ,

    so for the acetic acid it would be the same things for the moles right? because the NaOh doesn't change it is still 2.3 ml and then 8ml of acetic acid x .0008 L= .00023 mol also.

    then do we jus plug that into the Henderson equation?

    pH= pKa +log (NaOh)/(acetic acid)??

    how do we find pKa?

  • Chemistry -- to Dr.Bob - ,

    No, because sodium acetate is the salt of a strong base and a WEAK acid, (as opposed to HCl which is a strong acid) so you just follow the steps I did.
    write the equation.
    You have mols NaOH
    You have mols acetic acid.
    So how much sodium acetate (the salt) is produced? What is its concn? (mols/L)
    How much of the acetic acid is left unreacted? What is its concn? (mols/L)
    Those two (the salt is the base and the acetic acid unreacted is the acid) are the molarities you substitute into the H-H equation to obtain the pH of a buffer solution. The point of the whole exercise, I'm sure, is for you to see the DIFFERENCE in the pH between having an excess of a strong acid in solution with a neutral salt, such as NaCl, and having a weak acid/strong base salt, such as sodium acetate, along with its weak acid and forming a buffer solution. The pH of the strong acid, by itself is quite strong at 1.26. But the buffer solution is closer to 4 something. So you see the buffer is about 1000 times weaker AND it resists a change in pH (which is why we use buffers).

  • Chemistry -- to Dr.Bob - ,

    Ok so this is what i have so far...

    moles NaOh: 0.00023
    moles acetic acid: 0.008

    how much salt is produced: .008-.00023= .000777
    concentration of salt: .00777/.0103= .754

    acetic acid that is unreacted: .008- .00777=.0023
    concentration of acetic acid: .0023/.0103=.0223

    log (.754/.0223)= 1.53???

    and it is wrong. :(((((((((((((((((((((((

  • Chemistry -- to Dr.Bob - ,

    Ok so this is what i have so far...

    moles NaOh: 0.00023
    moles acetic acid: 0.008 I think 0.1 x 0.008 = 0.0008 mols acetic acid and that makes the remaining part of your problem incorrect; however, the procedure is correct except for the pKa bit. See below for pKa.

    how much salt is produced: .008-.00023= .000777 Wouldn't you have thought that the mols sodium acetate produced would be the same as the mols NaCl produced since it was the same two numbers subtracted?

    concentration of salt: .00777/.0103= .754 You didn't calculate a concn of the salt in the previous NaOH/HCl problem.

    acetic acid that is unreacted: .008- .00777=.0023 Mols of unreacted acid is the same as in the previous problem, I think.
    concentration of acetic acid: .0023/.0103=.0223

    log (.754/.0223)= 1.53??? You omitted pKa from the formula so you should expect to have a wrong answer!! right? The correct answer is, as I said in my earlier post, closer to 4. So what is pKa. Look in your text for Ka for acetic acid. Its about 1.8 x 10^-5 but use what you text or your teacher has provided. MOST modern texts now lists a pKa at the same time. If they don't do that, then you take the negative log of Ka to get pKa. That is pKa = -log Ka. My table lists 1.8 x 10^-5 for Ka, therefore, pKa = 4.74

    and it is wrong. :((((((((((((((((((((((( It won't be wrong if you do it right.;-)

  • Chemistry -- to Dr.Bob - ,

    wait.. im doing it again and getting this wrong.. just what did i need to correct again? are all my steps okay i just need to change my moles Acetic acid value? or is something wrong.

  • Chemistry -- to Dr.Bob - ,

    wait WAIT I GOT THE RIGHT ANSWER.. woooooooooooooooooooooooooooooo hoooooo thank you again.

  • Chemistry -- to Dr.Bob - ,

    You made a mistake in the mols acetic acid. M x L = 0.1 x 0.008 = 0.0008 so that made everything else wrong; however, I think the steps were ok.
    0.0008 mols acetic acid.
    0.1 M x 0.0023 L = 0.00023 mols NaOH.
    So you have formed 0.00023 mols salt.
    You have remaining 0.0008 - 0.00023 = 0.00057 mols acetic acid.
    Change to concns and use the HH equation. You should get pH about 4. Check my math.

  • Chemistry -- to Dr.Bob - ,

    yea i got 4.35 :)

  • Chemistry -- to Dr.Bob - ,

    Is it right? I'm going to bed if it is.

  • Chemistry -- to Dr.Bob - ,

    this is the second part of the same problem ...

    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 (like it will be in the experiment you perform in lab)?

    How do we go about this one? do we just multiply the .1 acid by 100 ml?

  • Chemistry -- to Dr.Bob - ,

    ya im exhausted too i went straight to theses questions right after school. are u gonna be on here tomorrow? cus i need help on like 3 more :)

  • Chemistry -- to Dr.Bob - ,

    whom should i link up with to meet the challenge of spending the next two hours blowing up rubber balloons by mouth besides old huff and puff luc,eh what?I`m woefully loneley for Mr.Adams.Thanks

  • Chemistry -- to Dr.Bob - ,

    At a certain temperature the following reactions have the constants shown:

    S(s) + O2(g) SO2(g0 k'c = 4.2 x 10^52
    2S(s) + 3O2(g) 2SO3(g) k"c = 9.8 x 10^128

    Calculate the equilibrium constant Kc for the following reaction at that temperture:

    2SO2(g) + O2(g) 2SO3(g)

    can you explain to me how to do this? I have the answer, but understand how they did it.

  • Chemistry -- to Dr.Bob - ,

    yes, I'll be on tomorrow.
    If you dilute the acid with 100 mL water, you don't change the mols of acid or the mols of base or the mols of salt formed or the mols of acid remaining.
    (a) In the NaOH/HCl problem, the only thing that changes is the final mols/L (mols is the same but L is different. If 100 mL water is added to what you have then the final volume is 100 mL + 2.3 mL + 8.00 mL = ??, that changes the final molarity of HCl) and that changes pH.
    b)In the NaOH/acetic acid problem, same thing except the M of the salt changes, the molarity of the acid changes, and that changes the HH equation.
    I did a quickie calcn (so don't hold me to it--I'm sleepy) and I get 2.29 for NaOH/HCl. And I get no change for the buffer. (Remember what I said about a buffer---it resists a change in pH so adding 100 mL water changes the first unbuffered problem but not the second buffered problem. Great things, these buffers! I'm outta here.

  • Chemistry -- to Dr.Bob - ,

    I know this convo happened a while ago lol. but i seriously don't know how to find the HCL pH for this:

    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

  • Chemistrylaboratory - ,

    (a)1. prepare a buffer solution by using acetic acid and sodium acetate.2. prepare M/40 NAOH solution.(b)1.aqueous solution of NAOHis electrolyte or non- electrolyte.2. measure the ph of the following :- lemom juice, coko -cola, NAOH SOLUTION

  • Chemistry laboratory - ,

    prepare a buffer solution by using acetic acid and sodium acetate.2. prepare M/40NAOH solution

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