## Chemistry: Molecular Approach (4th Edition)

$$[ClO^-] = 2.0 \times 10^{-7} \space M$$
In order to find the $ClO^-$ concentration, we have to calculate the ionization of the weakest acid, HClO. We can use the data from the example above to know the concentration of $H_3O^+$ at the start of the HClO ionization. 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HClO ]& [ ClO^- ]& [ H_3O^+ ]\\ Initial& 0.100 & 0 & 1.428 \times 10^{-2} \\ Change& -x& +x& +x\\ Equilibrium& 0.100 -x& 0 +x& 1.428 \times 10^{-2} +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ ClO^- ][ H_3O^+ ]}{[ HClO ]}$$ $$K_a = \frac{(x)( 1.428 \times 10^{-2} + x)}{[ HClO ]_{initial} - x}$$ 3. Assuming $0.100 \space and \space 1.428 x 10^{-2} \gt\gt x:$ $$K_a = \frac{(x)( 1.428 \times 10^{-2} )}{[ HClO ]_{initial}}$$ $$x = \frac{K_a \times [ HClO ]_{initial}}{ 1.428 \times 10^{-2} } = \frac{ 2.9 \times 10^{-8} \times 0.100 }{ 1.428 \times 10^{-2} }$$ $x = 2.0 \times 10^{-7}$ 4. Test if the assumption was correct: $$\frac{ 2.0 \times 10^{-7} }{ 0.100 } \times 100\% = 2.0 \times 10^{-4} \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 2.0 \times 10^{-7}$ 6. At equilibrium: $$[ClO^-] = x = 2.0 \times 10^{-7} \space M$$