## Chemistry: The Molecular Science (5th Edition)

The pH of blood containing these compounds is equal to $7.46$.
1. Drawing the ICE table, we get these concentrations at the equilibrium: $CH_3COOH(aq) + H_2O(l) \lt -- \gt CH_3C{OO}^-(aq) + H_3O^+(aq)$ Remember: Reactants at equilibrium = Initial Concentration - x And Products = Initial Concentration + x $[CH_3COOH] = 2 \times 10^{-3} M - x$ $[CH_3C{OO}^-] = 0.025M + x$ $[H_3O^+] = 0 + x$ 2. Calculate 'x' using the $K_a$ expression. $4.3\times 10^{- 7} = \frac{[CH_3C{OO}^-][H_3O^+]}{[CH_3COOH]}$ $4.3\times 10^{- 7} = \frac{( 0.025 + x )* x}{ 2\times 10^{- 3} - x}$ Considering 'x' has a very small value. $4.3\times 10^{- 7} = \frac{ 0.025 * x}{ 2\times 10^{- 3}}$ $4.3\times 10^{- 7} = 12.5x$ $\frac{ 4.3\times 10^{- 7}}{ 12.5} = x$ $x = 3.44\times 10^{- 8}$ Percent dissociation: $\frac{ 3.44\times 10^{- 8}}{ 2\times 10^{- 3}} \times 100\% = 1.72\times 10^{- 3}\%$ x = $[H_3O^+]$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 3.44 \times 10^{- 8})$ $pH = 7.46$