1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Solution's pH

  1. Mar 9, 2006 #1
    What is the pH of 340 mg of Ca(H2PO4)2 in 600 mL H20?

    Well, find M of Ca(H2PO4)2 by converting mg -> g -> mol and M = mol Ca(H2PO4)2/.6 L

    Is Ca(H2PO4)2 a buffer??? I don't know what to do. Do you do an equation of Ca(H2PO4)2 --> Ca (2+) + H2PO4 (-). What steps must I follow?

    Thank you for any tips.
     
  2. jcsd
  3. Mar 10, 2006 #2
    :cool: Easy, (as you said) first of all, (as you said) find the number of moles of Ca(H2PO4) then divide by .600 (600mL = 0.600L) to find the molarity. Remember molarity is moles/L.

    Buffers dont have anything at all to do with this problem. Now set up the equilibrium table like this:


    (umm...lets ignore the charges oh H2PO4 [-1] and HPO4 [-2])
    (lets also ignore the [H+] that HPO4 can contribute to the solution which is very small)
    (And finally, ignore Ca+2 from the reaction....since it is just an "spectator")

    Before reaction:

    H2P04_____________ <->_________H+___________HP04
    .600M__________________________0M________________0M


    After reaction:

    H2P04_____________ <->_________H+___________HP04
    .600M-X_____________________0M+X_____________0M+X



    In other words at equilibrium, concentrations will be:

    H2P04_____________ <->_________H+___________HP04
    .600M-X________________________X_______________X



    So, finally, set up the Equilibrium equation

    ka = [H+][A-]
    .........[HA]

    Where Ka is 6.2 * 10^-8 (you should have found that on your book somewhere...appendix maybe... or some table listing Ka and Kb values...or online)

    [H+] = X (Hydrogen ion concentration...see table above)
    [A-] = X (the conjugate base concentration in other words [HPO4])
    [HA] = 0.600M - X (initial concentration of acid)

    so:

    6.2 * 10^-8 = __(X)(X)___
    ......................0.600M-X



    Now solve for X...using many different ways (Quadratic equation, etc.)
    OR
    ignore the value of X being substracted and then:

    6.2 * 10^-8 = __(X)(X)___
    ......................0.600M


    X should be less than 5% of 0.600 (the famous "5% Rule"...in chemistry only I guess)

    solving the equation by ignoring X:
    X = 0.00019287...
    which is .0321% of 0.600M... obviously less than 5%
    Now,

    (remember X = [H+] = [HPO4] from table above)

    pH = -log(X)
    Which is the same thing as:
    pH = -log( [H+] )

    pH = -log(0.00019287)

    Finally,
    pH = 3.71
    :surprised

    By using Derive (math program) to solve the for X without ignoring any value, X = .00019284... almost no difference.

    So, there you got your answer, pH = 3.71

    Hope it helps,
    Haxx0rm4ster

    ___________________________
    "What luck for leaders that men do not think"
    -Adolf Hitler
     
    Last edited: Mar 10, 2006
  4. Mar 10, 2006 #3
    I thought you would have to figure out the molarity of the [itex]Ca(H_2PO_4)_2[/itex] and then multiply it by two to get the molarity of the [itex]H_2PO_4^-[/itex], and then do exactly what you said using 1.20 M instead of .600 M. I'm probably wrong though...

    I think you get a pH of 3.56 then.
     
  5. Mar 11, 2006 #4
    Wow, well, in fact, you made me notice I missed that point, but I didn't even mean to do exactly your problem since I didn't really use my time on finding the number of moles in .340g of Ca(H2PO4)2
    (I simply assumed to use 1 mole of Ca(H2PO4)2 [but it would have been wrong anyway since there are 2 moles of H2PO4 per Ca(H2PO4)2 ])
     
  6. Mar 11, 2006 #5

    GCT

    User Avatar
    Science Advisor
    Homework Helper

    Yeah, you're right.

    Haxx0rm4ster was right on track, except that if this is for analytical chemistry (quantitative analysis) you'll need to employ the ampholyte equation. In the case where you start off soley with the ampholyte you can employ pH=.5(pKa1+pKa2).
     
  7. Mar 11, 2006 #6

    Borek

    User Avatar

    Staff: Mentor

    ksinclair13: you are right about H2PO4- concentration, should be 2.906e-03M

    This is insane question. Correct answer is 4.87 and there is no way to calculate it. Well, no way to do it easily on paper.

    H2PO4- is amphiprotic substance. If concentrated enough, its pH is average of pKa1 and pKa2 (see explanation at pH of amphiprotic substance lecture - equation 12.11). However, this solution is too diluted and assumptions needed for the simplified solution are not obeyed. See table with the results of pH calculation of NaH2PO4 on the page mentioned.

    4.87 is a precise result that can be found using .

    Edit: GCT answered while I was editing my answer - he posted the same equation, but solution is too diluted for its use.
     
    Last edited: Mar 11, 2006
  8. Mar 12, 2006 #7

    GCT

    User Avatar
    Science Advisor
    Homework Helper

    If it is supposedly due dilute, soaring crane can employ the the more exact equation for an amphoteric substance. But from what I remember, as long as you start off with pure ampholyte, one can employ the average pKa equation.
     
  9. Mar 12, 2006 #8

    Borek

    User Avatar

    Staff: Mentor

    Please check the table I mentioned earlier. This equation holds only for solutions of at least 0.1M concentration (assuming "reasonable" values of both pKa constants). In the case of question posted by Soaring Crane difference between exact and approximate pH value is 0.2 unit, which is hardly acceptable.
     
  10. Mar 12, 2006 #9

    GCT

    User Avatar
    Science Advisor
    Homework Helper

    First off, the link you mentioned is from your site, so it's a bit redundant for me to observe the contents of that webpage and hearing your opinion on the same topic here, right?

    Another thing here, is that you're saying that this problem is impossible, which is not the case. We're talking about Soaring Crane's problem here, which is probably from quantitative analysis. So using the ampholyte equation or the more complicating equation should do, as I have stated.

    You may be right on the argument about dilution and the ampholyte equation, I'm going to check upon what you said, and get back to this topic later. Another thing, that I forgot to mention is that the Ka1 and Ka2 of the compound in question must be separated by a certain range, from what I remember, by a magnitude of 10^4.
     
  11. Mar 15, 2006 #10

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    I agree. I don't think this problem is unsolvable. I don't think you can use pH=.5(pKa1+pKa2), but the more exact equation can be used.

    Soaring Crane,
    Here H2PO4- can act both as an acid and a base.

    As an acid:
    H2PO4- + <===> H+ + HP042- ---k1

    As a base:

    H2PO4- + H2O <===> H3PO4 + OH- --- k2

    In fact, the HP042- formed can further ionize to give P043- and H+, but this can be ignored as this ionisation will be suppressed due to the common ion effect on H+.

    So, the net equation of the ampholyte H2PO4- where it acts as both an acid and a base is

    2 H2PO4- <===> H3PO4 + HP042-

    and the equilibirium constant for this reaction will be
    K=k2/k1

    If the initial concentration of H2PO4- is [itex]c[/itex], and the degree of dissociation is [itex] \alpha [/itex], then from the above equation, at equilibirium, the concentration of H2PO4- is [itex] c - 2c\alpha [/itex] and the concentrations of [H3PO4 and HP042- are [itex] c \alpha [/itex].

    [tex] K = \frac{c^2 \alpha^2}{c^2(1-2\alpha)^2} [/tex]

    [tex] K = \frac{\alpha^2}{(1-2\alpha)^2} = \frac{k_2}{k_1}[/tex]

    Since you know k2 and k1, you can solve for [itex] \alpha [/itex].

    Now to find the pH, plug this value of [itex] \alpha [/itex] into the equation where it acts as an acid you will be able to find the concentration of H+ and the pH

    Isn't this what you also mean GCT?
     
    Last edited: Mar 15, 2006
  12. Mar 15, 2006 #11

    Borek

    User Avatar

    Staff: Mentor

    I think you have made a mistake - assuming k1 and k2 are equilibrium constants for the first two reactions, K=k2*k1/Kw (neglecting water concentration). Please check your math.

    Can you show how you want to use [itex] \alpha [/itex] value as an acid? I don't understand the idea.
     
  13. Mar 15, 2006 #12

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    You are right Borek. That was a very serious error I made, and the analysis I posted after that is wrong.

    So, with a different approach, I think the right answer should be
    [H+]2 = k1k2[H2PO4-]/(k1 + [H2PO4-])

    and if, k1 <<< [H2PO4-], it reduces to pH=.5(pK1+pK2).
     
    Last edited: Mar 15, 2006
  14. Mar 15, 2006 #13

    GCT

    User Avatar
    Science Advisor
    Homework Helper

    yeah, I think you've got the right equations here, and as you've mentioned, with some assumptions the equation reduces to the average pKa equation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A Solution's pH
  1. PH of solution (Replies: 7)

  2. PH of solution (Replies: 1)

  3. PH of solution (Replies: 1)

  4. PH solution (Replies: 40)

Loading...