Chemistry

Phosphoric acid, H3PO4(aq), is a triprotic acid, meaning that one molecule of the acid has three acidic protons. Estimate the pH, and the concentrations of all species in a 0.300 M phosphoric acid solution.
pKa1: 2.16
pKa2: 7.21
pKa3: 12.32

Hint: Convert the pKa values to Ka values.

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  1. .......H3PO4 ==> H^++ H2PO4^-
    I........0.3......0......0
    C........-x.......x.......x
    E......0.3-x......x.......x

    k1 = (H^)(H2PO4^-)/(H3PO4)
    Substitute the E line into Ks expression and solve for x = (H^+), convert to pH.

    ........H2PO4^- ==> H^+ + HPO4^2-
    From k1 ionization you see(H+) = (H2PO4^-)
    k2 = (H^+)(HPO4^2-)/(H2PO4-)
    and since (H^+) is in the numerator and H2PO4^- is is the denominator, they cancel and k2 = (HPO4^2-)

    .......HPO4^2- ==> H^+ + PO4^3-

    k3 = (H^+)(PO4^3-)/(HPO4^2-)
    I would substitute (H^+) from the first calculation you did, (HPO4^-) from the second calculation, then solve for (PO4^3-).

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  2. How do you find the concentration of[OH-]? :C

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  3. Would it be
    pOH = 14 - pH
    Then, [OH-] = 10^-pOH
    After finding [OH-] find the mol ratio?

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  4. To find [OH]
    pOH =14 - pH
    [OH-] = 10^ - pOH

    mol ratio is not required

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  5. KA1= 6.9e-3
    KA2= 6.2e-8
    KA3= 4.8e-13
    KA2 and KA3 are so small, only use KA1 for [H+]

    H3PO4 -> H+ H2PO4
    I .350 0 0
    C -x +x +x
    E .35-x x x
    KA1= x^2/.35-x 6.9e-3
    x= .046 = [H+]
    pH= -log(.046)=1.34
    [OH-]=1e-14/.046
    =2.2e-13

    [H3PO4]=(.35-.046)M
    =.304M

    H2PO4- ->H+HPO42-
    I .046 .046 0
    C -y +y +y
    E .046-y .046-y y

    KA2= y(.046+y)/(.046-y)
    =6.2e-8
    y=6.2e-8

    [HPO4^2-]=6.2e-8M
    [H2PO4^-]=.046-6.2e-8=.046M

    HPO4^2- >H+PO4^3-
    I 6.2e-8 .046 0
    C -z +z +z
    E 6.2e-8-z .046+z z

    KA3= z(.0460/
    (6.2e-8)-z=4.8e-13

    z=6.5e-19M=
    [PO4^3-]

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  6. substitute your M for the M found here (0.350)

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