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Chemistry - Please help!

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What is the silver ion concentration in a solution prepared by mixing 433 mL of 0.356 M silver nitrate with 467 mL of 0.574 M sodium phosphate?
The Ksp of silver phosphate is 2.8 × 10^-18

  • Chemistry - Please help! -

    This is a limiting reagent, solubility product, and common ion effect problem all rolled into one.
    3AgNO3 + Na3PO4 ==> Ag3PO4 + 3NaNO3.

    mols AgNO3 = M x L = ? (apparox 0.15 but you should confirm the actual number.
    mols Na3PO4 = M x L = ? (about 0.27)

    Convert mols AgNO3 to mols of the product using the coefficients in the balanced equation.
    0.15 x (1/3) = about 0.05
    Convert mols Na3PO4 to mols Ag3PO4 = about 0.27

    In limiting reagent problems the SMALLER value ALWAYS wins; therefore, we have approximately 0.05 mols Ag3PO4 formed.
    How much Na3PO4 is used. That is
    0.15 mol AgNO3 x 1/3 = about 0.05 mols Na3PO4.
    How much Na3PO4 remains unreacted. That is 0.27-0.05 = about .22 mols. The concentration of the Na3PO4 is 0.22/(total volume in L) = ?M

    So we have a saturated solution of Ag3PO4 with an excess of ?M Na3PO4.

    ........Ag3PO4 ==> 3Ag^+ + PO4^3-
    ..........x.........3x......x
    Ksp = (Ag^+)^3(PO4^3-)
    For Ag substitute 3x
    For PO4^3- substitute x (for PO4 from Ag3PO4) + ?M from excess PO4^3- from Na3PO4. Solve for 3x = Ag^+

  • Chemistry - Please help! -

    Thank you so much!!!

  • Chemistry - Please help! -

    I am doing something wrong in my calculations.

    For ?M I have 0.24

    My equation is:
    2.8*10^(-18) = (3x)^3 * (x+0.24)

  • Chemistry - Please help! -

    I figured out what I was doing wrong! I have the right answer!
    Yay!

  • Chemistry - Please help! -

    Good for you.

  • Chemistry - Please help! -

    Hey i m not getting how you did the last part how did you solve for this equation
    2.8*10^(-18) = (3x)^3 * (x+0.24)
    what was your answer

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