Solution

Assume the label concentration as C and x mole ionized, then the ionization and the equilibrium concentrations can be represented by the ICE table below.

ICE (ce{CH3COOH}) (rightleftharpoons) (ce{CH_3COO^-}) (ce{H+}) Initial C 0 0 Change -x +x +x Equilibrium C – x x x

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with

[K_{ce a} = dfrac{x^2}{C – x} nonumber ]

The equation is then

[x^2 + K_{ce a} x = C K_{ce a} = 0 nonumber ]

The solution of x is then

[x = dfrac{- K_{ce a} + sqrt{K_{ce a}^2 + 4 C K_{ce a}}}{2} nonumber ]

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Recall that Ka – 1.78e-5, the values of x for various C are given below:

C = 1.0 0.010 0.00010 M x = 0.0042 0.00041 0.0000342 M pH = 2.38 3.39 4.47

DISCUSSION

In the above calculations, the following cases may be considered:

1. [x = sqrt{K_{ce a} times C} nonumber ] Note that you are comparing x with C here. If C > 100*Ka, the above method gives satisfactory results.
2. [x^2 + K_{ce a} x – C K_{ce a} = 0 nonumber ] and the solution for x, which must not be negative, has been given above.
3. Both cases 1 and 2 neglect the contribution of (ce{[H+]}) from the ionization of water. However, if the pH calculated from cases 1 and 2 falls in the range between 6 and 7, the concentration from self-ionization of water cannot be neglected. When the contribution of pH due to self-ionization of water cannot be neglected, there are two equilibria to be considered.

   (begin{array}{cccccl} ce{HA &rightleftharpoons &H+ &+ &A- &} C-x & &x & &x & ce{H2O &rightleftharpoons &H+ &+ &OH- &} 55.6 &&y &&y &leftarrow (ce{[H2O]} = 55.6) end{array})

   Thus,

   (begin{align} ce{[H+]} &= (x+y), ce{[A- ]} &= x, ce{[OH- ]} &= y, end{align})

   and the two equilibria are

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   [K_{ce a} = dfrac{(x+y) x}{C – x}label{1} ]

   and

   [begin{align} K_{ce w} &= (x+y), y, label{2} (K_{ce w} &= textrm{1E-14}) end{align} nonumber ]

   There are two unknown quantities, x and y in two equations, and (1) may be rearranged to give

   [x^2 + (y + K_{ce a}) x – C K_{ce a} = 0 nonumber ]

   [x = dfrac{ -(y+K_{ce a}) + ((y+K_{ce a})^2 + 4 C K_{ce a})^{1/2}}{2} nonumber ]

   One of the many methods to find a suitable solution for this problem is to use iterations, or successive approximations.

   1. Assume that (y = 1 10^{-7})
   2. [x = dfrac{ -(y+K_{ce a}) + ((y+K_{ce a})^2 + 4 CK_{ce a})^{1/2}}{2} nonumber ]
   3. [yn = dfrac{1 times 10^{-14}}{x+y} nonumber ]
   4. Replace y in step (2) by yn, and recalculate x.
   5. Repeat steps (2) and (3) until the new values and the old values differ insignificantly.

The above procedure is actually a general method that always gives a satisfactory solution. This technique has to be used to calculate the pH of dilute weak acid solutions. Further discussion is given in the Exact Calculation of pH.

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