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

The balanced chemical equation for the reaction is: \[ \mathrm{CH}_{3}\mathrm{COOH} + \mathrm{NaOH} \rightarrow \mathrm{CH}_{3}\mathrm{COO^{-}} + \mathrm{H}_{2}\mathrm{O} + \mathrm{Na^{+}} \]

To get the initial moles of acetic acid, use the given volume (35.0 mL) and concentration (0.150 M): Moles of acetic acid = volume × concentration Moles of acetic acid = \(35.0 \times 10^{-3} \mathrm{L} \times 0.150 \mathrm{M}\) Moles of acetic acid = \(5.25 \times 10^{-3}\) moles

For each volume of added NaOH, calculate the moles of NaOH, and then use the stoichiometry of the reaction to find the moles of acetic acid and acetate ion: (a) 0 mL of NaOH added: No reaction occurs, the solution remains as weak acetic acid. (b) 17.5 mL of NaOH added: Moles of NaOH = \(0.150 \mathrm{M} \times 17.5 \times 10^{-3} \mathrm{L}\) = \(2.625 \times 10^{-3}\) moles Since 1 mole of CH3COOH reacts with 1 mole of NaOH, the moles of CH3COOH remaining are \(5.25 \times 10^{-3} \mathrm{moles} - 2.625 \times 10^{-3} \mathrm{moles} = 2.625 \times 10^{-3} \mathrm{moles}\), and the moles of CH3COO- (acetate ion) produced are 2.625 × 10^{-3} moles. (c) - (f): Repeat the calculation for the other volumes of NaOH added.

Now we want to find the concentrations of acetic acid and the acetate ion at each stage of the titration. To do this, we divide the moles of each substance by the total volume of the solution at that point: Concentration of CH3COOH = moles of CH3COOH / total volume Concentration of CH3COO- = moles of CH3COO- / total volume For example, for 17.5 mL of NaOH added (b), the total volume is 52.5 mL (35.0 mL of acetic acid + 17.5 mL of NaOH): Concentration of CH3COOH = \(2.625 \times 10^{-3}\mathrm{moles} / (52.5 \times 10^{-3}\mathrm{L}) = 0.05\mathrm{M}\) Concentration of CH3COO- = \(2.625 \times 10^{-3}\mathrm{moles} / (52.5 \times 10^{-3}\mathrm{L}) = 0.05\mathrm{M}\) Repeat this process for the other volumes of NaOH added.

With the concentrations of acetic acid and acetate ion at each stage, we can now find the pH at each stage. For the stages before the equivalence point (when equal moles of acid and base have reacted), we can use the Henderson-Hasselbalch equation: pH = pKa + log(\(\dfrac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}\)) For CH3COOH, the pKa value is 4.74. For example, for 17.5 mL of NaOH added (b), we have: pH = 4.74 + log(\(\dfrac{0.05}{0.05}\)) = 4.74 For the stages after the equivalence point, we'll have excess NaOH. Therefore, we'll calculate the concentration of excess OH-, and then find the pOH. The pH can be found using the relationship: pH = 14 - pOH Repeat this process for the other volumes of NaOH added to find the pH at each stage.