## Chemistry 12th Edition

(a) $pH = 7.03$
$H_2{PO_4}^-(aq) + H_2O(l) \lt -- \gt HP{O_4}^{2-}(aq) + H_3O^+(aq)$ Remember: Reactants at equilibrium = Initial Concentration - x And Products = Initial Concentration + x $[H_2{PO_4}^-] = 0.15 M - x$ $[HP{O_4}^{2-}] = 0.10M + x$ $[H_3O^+] = 0 + x$ 2. Calculate 'x' using the $K_a$ expression. $6.2\times 10^{- 8} = \frac{[HP{O_4}^{2-}][H_3O^+]}{[H_2{PO_4}^-]}$ $6.2\times 10^{- 8} = \frac{( 0.1 + x )* x}{ 0.15 - x}$ Considering 'x' has a very small value. $6.2\times 10^{- 8} = \frac{ 0.1 * x}{ 0.15}$ $6.2\times 10^{- 8} = 0.67x$ $\frac{ 6.2\times 10^{- 8}}{ 0.67} = x$ $x = 9.3 \times 10^{- 8}$ Percent dissociation: $\frac{ 9\times 10^{- 8}}{ 0.15} \times 100\% = 6.2\times 10^{- 5}\%$ x = $[H_3O^+]$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 9.3 \times 10^{- 8})$ $pH = 7.03$