Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

pH question ?

A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. 

A solution is made by mixing 6.00 mmol (millimoles) of HA and 2.00 mmol of the strong base. What is the resulting pH?

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  • david
    Lv 7
    2 months ago

    NaOH is a strong base so I will use it.  Any other strong base would do the same.

       Ka = 5.61×10−6

      6.00 mmol HA  +  2.00 mmol NaOH ..  obviously the Acid is in excess  ....    4.00 mmol HA remain unreacted

      2.00mmol NaA are formed ...  The problem is that you gave NO VOLUMES  .....   the formula for Ka uses molar concentrations, not just moles. ===  U will use v for the unknown volume.

    HA  +  NaOH  -->  NaA  +  H2O

      NaA  -->  Na+  +  A-

        HA  <-->  H+  +  A-

     . . . . .  init . . . . . change . . . . . . equikub

      [HA] . . 4.00/v  . . - x/v . . . .   . .  4/v - x/v

      [H+] . .  0 . . . . . .  +x/v . . . . . . . .   x/v

      [A-] . .  2.00/v . . .  + x/v . . . . . . . 2/v + x/v

     = = = = = = = = = = = = = = = = = = = = = = = = 

      Ka = [H+][A-] / [HA]

     5.61x10^-6  =  (x/v)((4-x)/v) / ((2+x)/v)

     5.61x10^-6 = (x/v)((4-x)) / ((2+x))  <<<  2 unknowns x and v ... but only one equation  ....  impossible to solve.

      ...  assuming v = 1  ... then I'll continue

      5.61x10^-6 = (x)((4-x)) / (2+x)   <<<<<<   assume x is 'small' and round

      5.61x10^-6 = (x)((4)) / (2)  = 2x

       x = 2.805x10^-6  =  [H+]

      pH  =  -log 2.805x10^-6  =  5.55  <<< round as appropriate

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