## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 772: 71

#### Answer

The percent ionization is equal to 0.0063%

#### Work Step by Step

1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HCN ]& [ CN^- ]& [ H_3O^+ ]\\ Initial& 0.125 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.125 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CN^- ][ H_3O^+ ]}{[ HCN ]}$$ $$K_a = \frac{(x)(x)}{[ HCN ]_{initial} - x}$$ 3. Assuming $0.125 \gt\gt x:$ $$K_a = \frac{x^2}{[ HCN ]_{initial}}$$ $$x = \sqrt{K_a \times [ HCN ]_{initial}} = \sqrt{ 4.9 \times 10^{-10} \times 0.125 }$$ 4. Calculate the percent ionization: $$\frac{\sqrt{ 4.9 \times 10^{-10} \times 0.125 } }{ 0.125 } \times 100\% = 0.0063 \%$$ - Since the percent ionization is less than 5%, the assumption was valid.

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