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

The pH change is equal to $3.78$.
1. Find the concentration after the dilution: $C_1 * V_1 = C_2 * V_2$ $1*1= C_2 *100$ $1 = C_2 *100$ $C_2 =0.01$ 2. Drawing the ICE table we get these concentrations at the equilibrium: $CH_3COOH(aq) + H_2O(l) \lt -- \gt CH_3COO^-(aq) + H_3O^+(aq)$ Remember: Reactants at equilibrium = Initial Concentration - x And Products = Initial Concentration + x $[CH_3COOH] = 0.01 M - x$ $[CH_3COO^-] = 0.01M + x$ $[H_3O^+] = 0 + x$ 3. Calculate 'x' using the $K_a$ expression. $1.8\times 10^{- 5} = \frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$ $1.8\times 10^{- 5} = \frac{( 0.01 + x )* x}{ 0.01 - x}$ Considering 'x' has a very small value. $1.8\times 10^{- 5} = \frac{ 0.01 * x}{ 0.01}$ $1.8\times 10^{- 5} = 1x$ $\frac{ 1.8\times 10^{- 5}}{ 1} = x$ $x = 1.8\times 10^{- 5}$ Percent dissociation: $\frac{ 1.8\times 10^{- 5}}{ 0.01} \times 100\% = 0.18\%$ x = $[H_3O^+]$ $[CH_3COOH] = 0.01 M - x = 0.01 M - 1.8 \times 10^{-5}M \approx 0.01M$ $[CH_3COO^-] = 0.01M + x = 0.01 M + 1.8 \times 10^{-5}M \approx 0.01M$ $[H_3O^+] = 0 + x = 1.8 \times 10^{-5}M$ 4. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.8 \times 10^{- 5})$ $pH = 4.74$ 5. Since we are adding a strong base, this reaction will occur: $CH_3COOH(aq) + OH^-(aq) -- \gt CH_3COO^-(aq) + H_2O(l)$ And these are the concentrations after this reaction: Remember: Reactants at equilibrium = Initial Concentration - y And Products = Initial Concentration + y Since $NaOH$ is a strong base, y = $[NaOH] = 0.01M$ $[CH_3COOH] = 0.01 M - 0.01 = 0M$ $[CH_3COO^-] = 0.01M + 0.01 = 0.02M$ - The only compound that reacts in the solution is $CH_3COO^-$ with 0.02M So, calculate the pH of a solution with it. - Since $CH_3COO^-$ is the conjugate base of $CH_3COOH$ , we can calculate its kb by using this equation: $K_a * K_b = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_b = 10^{-14}$ $K_b = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_b = 5.6\times 10^{- 10}$ 6. We have these concentrations at equilibrium: -$[OH^-] = [CH_3COOH] = x$ -$[CH_3COO^-] = [CH_3COO^-]_{initial} - x = 0.02 - x$ For approximation, we consider: $[CH_3COO^-] = 0.02M$ 7. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][CH_3COOH]}{ [CH_3COO^-]}$ $Kb = 5.6 \times 10^{- 10}= \frac{x * x}{ 0.02}$ $Kb = 5.6 \times 10^{- 10}= \frac{x^2}{ 0.02}$ $1.1 \times 10^{- 11} = x^2$ $x = 3.3 \times 10^{- 6}$ Percent ionization: $\frac{ 3.3 \times 10^{- 6}}{ 0.02} \times 100\% = 0.017\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [CH_3COOH] = x = 3.3 \times 10^{- 6}M$ $[CH_3COO^-] \approx 0.02M$ 8. Calculate the pOH: $pOH = -log[OH^-]$ $pOH = -log( 3.3 \times 10^{- 6})$ $pOH = 5.48$ 9. Find the pH: $pH + pOH = 14$ $pH + 5.48 = 14$ $pH = 8.52$ 10. Calculate the pH change: $\Delta pH = 8.52 - 4.74 = 3.78$