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Calculating pH of the 1,96 % H3PO4 solution

  1. Jan 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate pH of the 1,96% solution of H3PO4 (ρ≈1 g/cm^3). Take in account only the first constant of dissociation (7,5 * 10^-3 mol/cubic dm). The reduction of the concentration of H3PO4 because of dissociation can be ignored.

    % = 1,96%
    ρ (H3PO4) ≈1 g/cm^3
    K = 7,5 * 10^-3 mol/cubic dm

    2. Relevant equations

    No equations in this task.

    3. The attempt at a solution

    Couldn't even attempt to solve it, because I don't know how to put the information together.
     
  2. jcsd
  3. Jan 13, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    Convert to the molar concentration first.

    You don't know anything about calculating pH of a weak acid? Equilibrium? Dissociation constants?
     
  4. Jan 14, 2012 #3
    I know a few formulas, but I think I can't use any of them in this task yet, because I don't have enough information.

    And which formula is used for molar concentration, this one:

    c (substance) = n (substance) / V (whole solution volume)

    or this one:

    b (substance) = n (substance) / m (mass of solvent)

    I know that pH can be calculated as

    pH = -log [H+] / mol dm^-3

    but that doesn't help at the moment. :/

    nor this one:

    [H+] = Kw / [OH-]

    I also know this:

    K = alpha^2 * c

    But none of this helps. :/
     
  5. Jan 14, 2012 #4

    Borek

    User Avatar

    Staff: Mentor

    There is enough information.

    First things first - you have to find a way to convert a %w/w concentration (given) to the molar concentration.

    Then it will be about dissociation. Can you write reaction equation for the first step of the phosphoric acid dissociation? Can you write formula for the equilibrium constant for this reaction? This will be identical to the definition of the dissociation constant Ka1 (given in the question, just marked K).
     
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