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Chapter 17: Problem 45

A 35.0-mL sample of 0.150\(M\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is titrated with 0.150 \(\mathrm{M} \mathrm{NaOH}\) solution. Calculate the pH after the following volumes of base have been added: (a) 0 \(\mathrm{mL}\) (b) \(17.5 \mathrm{mL},(\mathrm{c}) 34.5 \mathrm{mL},(\mathbf{d}) 35.0 \mathrm{mL},(\mathbf{e}) 35.5 \mathrm{mL},(\mathbf{f}) 50.0 \mathrm{mL}\)

Short Answer

Expert verified
The pH of the solution after various volumes of \(0.150 \ M \ NaOH\) is added to a \(35.0 \ mL\) sample of \(0.150 \ M \ CH_3COOH\) can be calculated using the acid dissociation constant and stoichiometry. The pH values are as follows: (a) after \(0 \ mL\) of NaOH, pH = \(2.86\); (b) after \(17.5 \ mL\) of NaOH, pH = \(4.74\); (c) after \(34.5 \ mL\) of NaOH, pH = \(7.95); (d) after \(35.0 \ mL\) of NaOH, pH = \(8.82\); (e) after \(35.5 \ mL\) of NaOH, pH = \(9.23\); (f) after \(50.0 \ mL\) of NaOH, pH = \(10.21\).

Step by step solution

01

The balanced chemical equation for the reaction of acetic acid (CH3COOH) and sodium hydroxide (NaOH) is: \( CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O \) #Step 2: Calculate the initial moles of acetic acid and moles of NaOH added#

The initial moles of acetic acid can be calculated using the given concentration and volume of the solution: Initial moles of \(CH_3COOH =\) Concentration × Volume Initial moles of \(CH_3COOH = 0.150 M \times 0.035 L = 0.00525 mol\) At each step, calculate the moles of NaOH added using the given volumes: Moles of \(NaOH_{added} =\) Concentration × Volume We will do this for each volume given in the problem (0 mL, 17.5 mL, 34.5 mL, 35.0 mL, 35.5 mL, and 50.0 mL). #Step 3: Determine the moles of the remaining acetic acid, moles of the produced acetate ion, and moles of the remaining NaOH#
02

For each volume of added NaOH, calculate the moles of the remaining acetic acid, the moles of the produced acetate ion, and the moles of the remaining NaOH using stoichiometry and an ice table: Moles of \(CH_3COOH_{remaining} =\) Initial moles of \(CH_3COOH -\) Moles of \(NaOH_{added}\) (if NaOH is limiting reagent) Moles of \(CH_3COO^-\) = Moles of \(NaOH_{added}\) (if NaOH is limiting reagent) Moles of remaining \(NaOH =\) Moles of \(NaOH_{added} -\) Initial moles of \(CH_3COOH\) (if acetic acid is limiting reagent) Calculate these values for each given volume of added NaOH. #Step 4: Calculate the pH of the solution using the ion concentrations and the acid dissociation constant of acetic acid#

In order to calculate the pH of the solution, we need to use the dissociation constant of acetic acid, which is indicated as \(K_a\): \( K_a = 1.8 \times 10^{-5} \) For each given volume, we will either have: 1) Excess acetic acid: Use the Henderson-Hasselbalch equation to calculate the pH when there is acetic acid and acetate ion: \(pH = pK_a + \log \frac{[CH_3COO^-]}{[CH_3COOH]}\) 2) Excess NaOH: Calculate the concentration of the remaining NaOH in the solution, and then use the negative logarithm to determine the pH of the solution: \[pH = -\log([OH^-])\] Perform these calculations for each volume to find the pH of the solution at each stage of the titration.

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Most popular questions from this chapter

Predict whether the equivalence point of each of the following titrations is below, above, or at \(\mathrm{pH} 7 :\) (a) formic acid titrated with \(\mathrm{NaOH},(\mathbf{b})\) calcium hydroxide titrated with perchloric acid, (c) pyridine titrated with nitric acid.

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