120 mL of 1.45 M sodium dihydrogen phosphate (NaH2PO4) is mixed with 280 mL of 0.987 M sodium hydrogen phosphate (Na2HPO4).?

a. What is the original pH of this buffer?

b. What is the pH of the buffer if 1.00 mL of 12 M HCl is added to the buffer from

part (a)?

c. What is the pH of the buffer is 1.00 mL Of 18 M NaOH is added to the buffer

from part (a)?

1 Answer

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  • david
    Lv 7
    4 months ago

    a,   Ka for NaH2PO4  =  7.586x10^-3

    Ka for Na2HPO4  =  ??  can't find this with GOOGLE .. but it is basic, so I will assume it is small enough to ignore

    120 mL of 1.45 M sodium dihydrogen phosphate (NaH2PO4) is mixed with 280 mL of 0.987 M sodium hydrogen phosphate (Na2HPO4).

      H2PO4(-)  <-->  HPO4(2-)  +  H+

       total vol  =  400mL  ---   initial [H2PO4(-)]  =  1.45 X 120/400 

        init [H2PO4(-)] = 0.435 M

      init [HPO4(2-)]  =  0.987 X 280/400  =  0.6909 M

       Ka  =  [HPO4(2-)][H+] / [H2PO4(-)]

      7.586x10^-3  =  (0.6909 + x)(x) / (0.435 - x)  <<<  if 'x' is small, round

      7.586x10^-3 = (0.6909)(x) / (0.435)   <<<  solve

       x =  0.004776 M  <<<  see, x is small

       pH  =  2.32  <<<  round as needed  answer for a.

    b.  12.0M  X  1.0/1000  =  0.0120 mole H+ is added  --  this reacts with the buffer in reverse ..  using HPO4(2-) and forming more H2PO4(-)

       total vol NOW  =  401 mL

      

       init [HPO4(2-)] =  0.6909 - 0.0120/0.401  =  0.661 M

      init [H2PO4(-)] = 0.435 M + 0.0120/0.401 =  0.464925 M

        Ka = [HPO4(2-)][H+] / [H2PO4(-)]

      7.586x10^-3 = (0.661 + x)(x) / (0.464925 - x) <<< if 'x' is small, round

       7.586x10^-3 = (0.661)(x) / (0.464925)  <<<  slve for x

       x  =  0.0053357 M  << = [H+]

       pH  =  2.273  <<<  rounnd as needed

    now you try part c

      

        

      

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