? 發問於 科學及數學化學 · 1 十年前

About Buffer and Ksp....how to calculate??thx!!

1) What is the maximum amount of sodium sulfate that can be added to 1.00 L of 0.0020 Ca(NO3)2 before precipitation of calcium sulfate begins? Ksp = 2.4 × 10-5 for calcium sulfate

2) A buffer is prepared by adding 150 mL of 1.0 M NaOH to 250 mL of 1.0 M NaH2PO4. How many moles of HCl must be added to this buffer solution to change the pH by 0.18 units?

更新:

Thank you for answering my question.

but could you explain why ''Since 1 mole of sodium sulphate contains 2 moles of SO42- ions, the maximum allowable amount of sodium sulphate to be added into the solution is 0.006 moles, or, alternatively, 0.852 g.'' in question 1?

Thx a lot!!

1 個解答

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  • 1 十年前
    最愛解答

    (1) First of all, concenctration of Ca2+ ions = 0.002 M

    Therefore, maximum allowable concentration of SO42- ions for no precipitation formed is:

    2.4 × 10-5/0.002 = 0.012 M

    Since 1 mole of sodium sulphate contains 2 moles of SO42- ions, the maximum allowable amount of sodium sulphate to be added into the solution is 0.006 moles, or, alternatively, 0.852 g.

    (2) Initially, the no. of moles of the substances are given below:

    NaOH = 0.15 moles

    NaH2PO4 = 0.25 moles

    Then, upon mixing, all NaOH are consumed to give Na2HPO4 and therefore the no. of moles of the product is 0.15 moles too while 0.1 moles of NaH2PO4 are left.

    So after mixing the solutions, we have:

    Concentration of HPO42- ions = 0.15/0.4 = 0.375 M

    Concentration of H2PO4- ions = 0.1/0.4 = 0.25 M

    And, from the deductions below:

    圖片參考:http://i117.photobucket.com/albums/o61/billy_hywun...

    where Ka2 is the second acid dissociation constant of phosphoric acid.

    Suppose that, to decrease the pH value by 0.18, the change in concentration of HPO42- ions and H2PO4- ions are -x and +x respectively, we have:

    圖片參考:http://i117.photobucket.com/albums/o61/billy_hywun...

    So a change in concentration of 0.0639 M is needed for each species and hence, an addition of 0.0639 M of HCl is also needed since the conversion between the species needs 1 mole of HCl for each mole of conversion.

    Finally, the no. of moles of HCl needed is:

    0.0639 × 0.4 = 0.0256 moles

    2007-06-14 00:12:53 補充:

    I am so sorry for my serious mistake:It should be "Since 1 mole of sodium sulphate contains 1 mole of sulphate ions"Then the maximum allowable concentration of sodium sulphate will also be 0.012M.So to speak, to 1.00 L, 0.012 moles of sodium sulphate= 1.704 g can be added in maximum.

    資料來源: My chemical knowledge
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