Equilibrium expression quadratic equation solution

The solution is that of a quadratic equation. If less than about 10 or 15% of the acid is ionized, the expression may be simphfied by neglecting x compared with C (10 M in this case). This is an arbitrary (and not very demanding) criterion. The simplification applies if Ka is smaller than, about 0.01C, that is, smaller than 10-" at C = 0.01 M, 10 at C = 0.1 M, and so forth. Under these conditions, the error in calculation is 5% or less (results come out too high), and within the probable accuracy of the equilibrium constant. Our calculation simplifies to... 

A common application of a quadratic equation in elementary chemistry is the calculation of the hydrogen ion concentration in a solution of a weak acid. If activity coefficients are assumed to equal unity, the equilibrium expression in terms of molar concentrations is... 

The general approach we have put forward in these example problems will work for any equilibrium calculation. But we should point out that these examples were chosen at least in part because they can be solved with relative ease. Even our more complicated case required nothing more than using the quadratic formula. But not all equilibria lend themselves to such simple solutions. Consider the An-drussow process for the production of HCN. The overall equation is shown below. Because all species are gases under the high temperature conditions needed for reaction, five concentration terms appear in the equilibrium expression, along with exponents as high as six ... 

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