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

Published by Cengage Learning

# Chapter 15 - Additional Aqueous Equilibria - Questions for Review and Thought - Topical Questions - Page 693a: 29a

#### Answer

The pH change is equal to $0.08$. ** Considering only one significant digit in the pH: The pH change is equal to $0.1$

#### Work Step by Step

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 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 $[CH_3COOH] = 0.1 M - x$ $[CH_3COO^-] = 0.1M + x$ $[H_3O^+] = 0 + x$ 3. Calculate 'x' using the $K_a$ expression. $1.8\times 10^{- 5} = \frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$ $1.8\times 10^{- 5} = \frac{( 0.1 + x )* x}{ 0.1 - x}$ Considering 'x' has a very small value. $1.8\times 10^{- 5} = \frac{ 0.1 * x}{ 0.1}$ $1.8\times 10^{- 5} = 1x$ $\frac{ 1.8\times 10^{- 5}}{ 1} = x$ $x = 1.8\times 10^{- 5}$ Percent dissociation: $\frac{ 1.8\times 10^{- 5}}{ 0.1} \times 100\% = 0.018\%$ x = $[H_3O^+]$ $[CH_3COOH] = 0.1 M - x = 0.1 M - 1.8 \times 10^{-5}M \approx 0.1M$ $[CH_3COO^-] = 0.1M + x = 0.1 M + 1.8 \times 10^{-5}M \approx 0.1M$ $[H_3O^+] = 0 + x = 1.8 \times 10^{-5}M$ 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 = $[NaOH] = 0.01M$ $[CH_3COOH] = 0.1 M - 0.01 = 0.09M$ $[CH_3COO^-] = 0.1M + 0.01 = 0.11M$ 6. Now, calculate the hydronium ion concentration after the addition of the $NaOH$: $[H_3O^+] = Ka * (\frac{[CH_3COOH]}{[CH_3COO^-]})$ $[H_3O^+] = 1.8 \times 10^{-5} * \frac{0.09}{0.11}$ $[H_3O^+] = 1.8 \times 10^{-5} * 0.82$ $[H_3O^+] = 1.5 \times 10^{-5}$ 7. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.5 \times 10^{- 5})$ $pH = 4.82$ 8. Find the pH change: $\Delta pH = 4.82 - 4.74 = 0.08$ ** Considering one significant digit in pH: $\Delta pH = 4.8 - 4.7 = 0.1$

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