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Chapter 15: Problem 6

Sketch a pH curve for the titration of a weak acid (HA) with a strong base (NaOH). List the major species, and explain how you would go about calculating the pH of the solution at various points, including the halfway point and the equivalence point

Short Answer

Expert verified
A pH curve for the titration of a weak acid (HA) with a strong base (NaOH) can be sketched by calculating the pH at various points throughout the titration, such as the initial point, halfway point, and equivalence point. The major species participating in the reaction are HA, OH-, A-, and H2O. To calculate the pH at different points, use the Ka for the weak acid (HA), the Henderson-Hasselbalch equation at the halfway point, and the Kb for the conjugate base (A-) at the equivalence point. Plot the calculated pH values on a graph with volume of NaOH on the x-axis and pH on the y-axis. The curve will show an initial slow increase in pH, a rapid increase near the equivalence point, and a slower increase after the equivalence point.

Step by step solution

01

Understand the Reaction and Major Species Present

For the titration of a weak acid (HA) with a strong base (NaOH), the reaction that is taking place is: HA + OH- → A- + H2O (as Na+ is not involved in the acidic or basic behavior of the solution) Where HA represents the weak acid, OH- represents the hydroxide ion from the strong base, A- represents the conjugate base of the weak acid, and H2O represents water. During the titration process, the major species present will change based on the amount of base being added.
02

Calculate pH at the Beginning of the Titration

Before any base is added, only the weak acid, HA, is present in the solution. To calculate the initial pH, we have to use the acid dissociation constant (Ka) for the weak acid as follows: \(K_a = \frac{[H^+][A^-]}{[HA]}\) At the beginning, the concentration of HA can be calculated by \(C_{HA} = \frac{mols}{volume}\). Since \( [A^{-}]=0 \) when no base is added, the equation becomes: \(K_a = \frac{x^2}{C_{ha} - x}\) Solve for x, which represents the concentration of H+. Then calculate the initial pH using the formula: pH = -log10([H+])
03

Calculate pH at the Halfway Point

At the halfway point of the titration, exactly half of the weak acid has been neutralized by the strong base, meaning that the concentration of the conjugate base (A-) is equal to the concentration of the remaining weak acid (HA). At this point, the Henderson-Hasselbalch equation can be used to find the pH: pH = pKa + log(\(\frac{[A^{-}]}{[HA]}\)) Since \([\text{A}^-\]=[\text{HA}]\) at the halfway point, the equation becomes pH = pKa.
04

Calculate pH at the Equivalence Point

At the equivalence point, all of the weak acid has been neutralized by the strong base, and only the conjugate base (A-) is left. Since A- is a weak base, we must now consider its hydrolysis with water: A- + H2O → HA + OH- To determine the concentration of OH-, we can use the base dissociation constant (Kb): \(K_b = \frac{[HA][OH^-]}{([A^{-}])}\) Note that Ka x Kb = Kw = 1 x 10^(-14). Solve for [OH-] to calculate the pOH and then find the pH: pH = 14 - pOH
05

Sketch the pH Curve

Using the pH values calculated at the initial, halfway, and equivalence points, plot these points on a graph, with the x-axis representing the volume of NaOH added and the y-axis representing the pH. Sketch a curve that connects these points, showing an initial increase in pH at a slow rate, followed by a rapid increase in pH around the equivalence point, and finally, a slower increase in pH as more base is added after the equivalence point.

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Most popular questions from this chapter

One method for determining the purity of aspirin $\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)$ is to hydrolyze it with NaOH solution and then to titrate the remaining NaOH. The reaction of aspirin with NaOH is as follows: \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}(s)+2 \mathrm{OH}^{-}(a q)\) $$ \mathrm{C}_{7} \mathrm{H}_{3} \mathrm{O}_{3}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ A sample of aspirin with a mass of 1.427 g was boiled in 50.00 \(\mathrm{mL}\) of 0.500\(M \mathrm{NaOH}\) . After the solution was cooled, it took 31.92 \(\mathrm{mL}\) of 0.289 \(\mathrm{M} \mathrm{HCl}\) to titrate the excess NaOH. Calculate the purity of the aspirin. What indicator should be used for this titration? Why?

Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A 25.0 -mL sample of 0.100\(M\) lactic actid $\left(\mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right.\( \)\mathrm{p} K_{\mathrm{a}}=3.86 )$ is titrated with 0.100 \(\mathrm{M}\) NaOH solution. Calculate the pH after the addition of $0.0 \mathrm{mL}, 4.0 \mathrm{mL}, 8.0 \mathrm{mL}, 12.5 \mathrm{mL},\( \)20.0 \mathrm{mL}, 24.0 \mathrm{mL}, 24.5 \mathrm{mL}, 24.9 \mathrm{mL}, 25.0 \mathrm{mL}, 25.1 \mathrm{mL}\( \)26.0 \mathrm{mL}, 28.0 \mathrm{mL}$ , and 30.0 \(\mathrm{mL}\) of the NaOH. Plot the results of your calculations as \(\mathrm{pH}\) versus milliliters of NaOH added.

The active ingredient in aspirin is acetylsalicylic acid. A 2.51 -g sample of acetylsalicylic acid required 27.36 \(\mathrm{mL}\) of 0.5106 $\mathrm{M} \mathrm{daOH}\( for complete reaction. Addition of 13.68 \)\mathrm{mL}$ of 0.5106\(M \mathrm{HCl}\) to the flask containing the aspirin and the sodium hydroxide produced a mixture with pH \(=3.48 .\) Determine the molar mass of acetylsalicylic acid and its \(K_{2}\) value. State any assumptions you must make to reach your answer.

A certain acetic acid solution has \(\mathrm{pH}=2.68\) . Calculate the volume of 0.0975 \(\mathrm{M} \mathrm{KOH}\) required to reach the equivalence point in the titration of 25.0 \(\mathrm{mL}\) of the acetic acid solution.

Calculate the ph of each of the following solutions. $$ \begin{array}{l}{\text { a. } 0.100 M \text { propanoic acid }\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{2}=1.3 \times 10^{-5}\right)} \\ {\text { b. } 0.100 M \text { sodium propanoate }\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)} \\ {\text { c. pure } \mathrm{H}_{2} \mathrm{O}}\end{array} $$ $$ \begin{array}{l}{\text { d. a mixture containing } 0.100 M \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} \text { and } 0.100 \mathrm{M}} \\\ {\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}}\end{array} $$
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