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chemistry

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I have many questions. I have a test tuesday and need to get these problems done and understand them today because i am currently studying. I am in college and my professor has not emailed me back regarding solutions for the review problems she posted. Is it possible if you can help me with these questions.

1) If Ka is small, concentration of w acid (HA) are large. What is a quick way of calculation (H+).
a)(H+)=Ka
b)(H+)=Ka (Ha)
c)(H+)=(Ka(Ha)
d)(H+)=KaKb(hA)

2) Which of the following salts forms aqueous solutions with pH=7? (How am I suppose to know this from these answers)

a)Na2S b)NaBR c) NaCLO2 d)NaNO2



3)What is the appro pH of solution of NaF?
well the answer is above 7, flouride is a weak base, (but i thought when added with water they neutralize each other forming ions).


4) What is the pH of a 0.15M solution of ammonium bromide? The kb value is 1.8E-5.
(How do i set this up, i tried doing it in place of Ka but that did not work).


5) What is the pH of a 100 ml solution made from 46.2 ml of ).050 M nitric acid and water?

I did stoichometry to try and figure this out and then plug into pH equation but it did not work, what did i do wrong.



6) A buffer system is set up with [HA]=2[A^-]. If Pka is 5.5 what is the pH of the buffer?




7) pH of pure water at 70C knowing pKw=13.16?

I know that at 20C pH=7.


i know these are a lot of questions, but I need the help to set these up and understand what I am doing thank you.

  • chemistry -

    I'm an idiot for starting a bunch of questions this long BUT you sounded desperate. Post separately if you have questions about my work.
    1) If Ka is small, concentration of w acid (HA) are large. What is a quick way of calculation (H+).
    a)(H+)=Ka
    b)(H+)=Ka (Ha)
    c)(H+)=(Ka(Ha)
    d)(H+)=KaKb(hA)

    I would go with a. b and c appear to be the same. I assume this is a PURE weak acid and not a buffer. Here is my method for a quick ESTIMATE.
    HA ==> H^+ + A^-
    Ka = (H^+)(A^-)/(HA) and
    (H^+) = sqrt[Ka*(HA)].
    I assumed Ka = 1E-10 and (HA) = 10 so (H^+) = 3.16E-5 as the correct answer.
    a)gives you 1E-5; close.
    b)gives you 1E-10 x 10; not close.
    c)give you the same; not close.
    d)gives you 1E-14*(10); not close.


    2) Which of the following salts forms aqueous solutions with pH=7? (How am I suppose to know this from these answers)
    a)Na2S b)NaBR c) NaCLO2 d)NaNO2
    You go with the hydrolysis of the salt. The chemical way of doing the first one is to say that Na^+ is not hydrolyzed because it is a weak acid but S^2- is hydrolyzed because it is a weak base. The hydrolysis is
    S^= + HOH ==> HS^- + OH^- and you can see that the excess OH^- will make the solution basic. However, the EASY way to do it is this.
    Na2S + HOH ==> NaHS + NaOH
    The products are a salt (an acid salt, to be sure, but still a salt and a weak acid) plus a strong base, NaOH. So the solution is basic.
    Second one is NaBr so the products are
    NaBr + HOH ==> HBr + NaOH
    Strong acid, strong base, solution of NaBr will be neutral.
    The other two are done the same way but you should have basic for NaNO2 and basic for NaClO2.



    3)What is the appro pH of solution of NaF?
    well the answer is above 7, flouride is a weak base, (but i thought when added with water they neutralize each other forming ions).
    You are right that F^- is a weak base.
    Kb = (Kw/Ka) = about (1#-14/1E-3) = about 1E-11. Then
    F^- + HOH ==> HF + OH^-
    Kb = 1E-11 = (OH^-)^2/(0.1) = about 1E-6 for pOH = 6 and pH about 8. I just made up the 0.1M for (F^-) because 0.1M solns are common in problems. You know Kw and you can guess Ka for HF is approx 1E-3.


    4) What is the pH of a 0.15M solution of ammonium bromide? The kb value is 1.8E-5.
    (How do i set this up, i tried doing it in place of Ka but that did not work).
    NH4^+ + H2O ==> NH3 + H3O^+
    Ka = (Kw/Kb) = (NH3)(H3O^+)/(NH4^+)
    (NH3) = x = (H3O^+), YOu know Kw and Kb and (NH4^+) 0.15-x which can be calcuated as 0.15 since x is small.


    5) What is the pH of a 100 ml solution made from 46.2 ml of ).050 M nitric acid and water?
    HNO3 is a strong acid and ionizes 100%. I suppose the problem means we use 46.2 mL of 0.050 M HNO3 and add enough water to make 100 mL total volume. Then (HNO3) = 0.050M x (46.2/100) = (H^+) and convert that to pH.


    I did stoichometry to try and figure this out and then plug into pH equation but it did not work, what did i do wrong.



    6) A buffer system is set up with [HA]=2[A^-]. If Pka is 5.5 what is the pH of the buffer?
    Use the Henderson-Hasselbalch equation.
    pH = pKa + log (base)/(acid)
    pH = 5.5 + log(A^-)/(2A^-) and solve for pH. The answer is about 5.2


    7) pH of pure water at 70C knowing pKw=13.16?

    I know that at 20C pH=7.
    H2O ==> H^+ + OH^-
    The ion product of water is (H^+)(OH^-) and pH + pOH = pKw. SO, if pKw = 13.16, then pH must be 1/2 (13.16) = 6.58 and pOH must be 1/2(13.16)= 6.58 and guess what? The solution is neutral since pH = pOH. The trick part of the question is that students get all caught up in the 7.0 being the neutral point and they don't get past the point that pH = pOH so it must be neutral. It is still neutral BECAUSE the pKw is not 14 but something different. If pKw = 10 (not likely), then the pH would be 5 and the pOH would be 5 BUT THE SOLUTION (water) WOULD BE NEUTRAL.



    i know these are a lot of questions, but I need the help to set these up and understand what I am doing thank you.

  • chemistry --TYPO -

    On question 2, I wrote the Na^+ was a weak acid; it is not.
    ON question 3, line 2, I wrote 1#-14 which should have been 1E-14. That's all I saw but I may have missed some typos.

  • chemistry -

    for #1:i still don't get it because the answer is c) [H+]=(Ka[HA]).

  • chemistry -

    For #1, what is the difference between answer (b) and answer (c)? They look the same to me. And (c) is correct only if you made a typo and the answer (c) is (H^+) = sqrt[Ka(HA)]. If that is the case I don't see that as a quick and dirty procedure for calculating (H^+). It's the standard way.

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