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A weak monoprotic acid is titrated with \(0.100 \mathrm{M} \mathrm{NaOH}\). It requires \(50.0 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the equivalence point. After \(25.0 \mathrm{~mL}\) of base is added, the pH of the solution is \(3.62\). Estimate the pKa of the weak acid.

Short Answer

Expert verified
The estimated pKa of the weak monoprotic acid is approximately 18.67.

Step by step solution

01

Calculate moles of NaOH required to reach the equivalence point

The molarity (M) is given as moles/volume. Rearrange the formula to calculate moles: Moles = Molarity × Volume Since the equivalence point is reached after adding 25.0 mL of 0.100 M NaOH: Moles of NaOH = 0.100 M × 0.025 L = 0.00250 moles
02

Find the concentration of the weak acid

In a titration, equivalence point is reached when the number of moles of acid equals the number of moles of base. Therefore, the moles of the weak monoprotic acid would also be 0.00250 moles. We know that during titration: Moles of acid = moles of base Since the volume of NaOH to reach the equivalence point is 25.0 mL, the volume of the weak acid is also 25.0 mL. To find the concentration of the weak acid, we will use the formula: Molarity = Moles/Volume Weak acid concentration = 0.00250 moles / 0.025 L = 0.100 M
03

Calculate the concentration of the dissociated components

Suppose the equilibrium concentrations in the solution after 12.5 mL of 0.100 M NaOH is added are as follows: - [HA] = 0.100 - x - [A-] = x - [OH-] = x At this point, HA has reacted with OH⁻ to form A⁻ and H₂O. Given that the pH of the solution is 4.16, we can find the concentration of [OH⁻] using: pOH = 14 - pH pOH = 14 - 4.16 = 9.84 Now, we can calculate the concentration of OH⁻ ions using the formula: [OH⁻] = 10^(-pOH) [OH⁻] = 10^(-9.84) = 1.45 × 10⁻¹⁰ M (approximately) This concentration is equal to x.
04

Calculate Ka of the weak acid

Now we can use the equilibrium concentrations to calculate the Ka of the weak acid. Ka = ([A⁻][H⁺])/[HA] Since [A⁻] = [OH⁻] and [H⁺] = [OH⁻], using the values of [OH⁻] obtained in step 3: Ka = (x^2)/(0.100 - x) We can assume that x is very small and therefore (0.100 - x) ≈ 0.100. Now, we can substitute the value of x: Ka = (1.45 × 10⁻¹⁰)^2 / (0.100) ≈ 2.11 × 10⁻¹⁹
05

Calculate pKa of the weak acid

Finally, we can find the pKa by taking the negative logarithm of Ka: pKa = -log(Ka) pKa = -log(2.11 × 10⁻¹⁹) ≈ 18.67 The estimated pKa of the weak monoprotic acid is approximately 18.67.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Weak Acid
A weak acid is a type of acid that doesn't completely ionize in water. This means that, when dissolved, only a small portion of its molecules release hydrogen ions (H⁺) into the solution. In chemical terms, it establishes an equilibrium between the undissociated acid (\(HA\)) and the dissociated ions (\([H^+]\) and \([A^-]\)).
Here are some characteristics of weak acids:
  • They have higher pKa values compared to strong acids.
  • Weak acids are less conductive than strong acids due to fewer ions in solution.
  • They react slower with bases during titration compared to strong acids.
Understanding the behavior of weak acids is crucial in titration since it affects the determination of various properties such as pKa, equivalence point, and more.
pKa
The pKa is a measure of the strength of an acid. It is the negative logarithm of the acid's dissociation constant (\(Ka\)). The smaller the pKa value, the stronger the acid because it dissociates more in water.
To calculate the pKa:
  • First, find the \(Ka\) of the acid, which shows how completely the acid dissociates into its ions.
  • Then, use the formula: \(pKa = -\log(Ka)\).
In titrations, knowing the pKa helps in identifying the acid and understanding its reactivity with bases. It is especially useful when plotting titration curves, where the midpoint of the buffer region equals the pKa of the weak acid.
Equivalence Point
The equivalence point in a titration is reached when the amount of titrant added equals the number of moles of the substance being titrated. For an acid-base titration, this is where the moles of acid are equal to the moles of base added. At the equivalence point:
  • The solution typically undergoes dramatic changes in pH.
  • It is not necessarily at a neutral pH, especially in the titration of a weak acid with a strong base.
  • Accurate determination requires indicators that change color at the right pH or precise pH meters.
Understanding the equivalence point is vital as it allows for the calculation of the concentration of the unknown solution, enhancing our understanding of that acid or base.
Molarity
Molarity (M) is a way to measure the concentration of a solution, specifically showing how many moles of solute are present in one liter of solution. It's crucial in titration calculations because it helps identify the exact amounts of reactants that will completely react with each other.
To calculate molarity:
  • Use the formula: \(Molarity = \frac{moles\ of\ solute}{liters\ of\ solution}\).
  • Knowing the molarity allows for the prediction of reactions and understanding the reactant proportions in solutions.
  • Titration often involves standard solutions of known molarity to determine the concentration of unknown solutions through calculation.
Dissociation Constant
The dissociation constant (\(Ka\)) is an equilibrium constant that measures the tendency of a weak acid to dissociate into its ions in water. It's a quantitative way to express the strength of a weak acid or base.
The formula for the dissociation constant is:
  • \(Ka = \frac{[H^+][A^-]}{[HA]}\), where \([H^+]\) and \([A^-]\) are the concentrations of the ions, and \([HA]\) is the concentration of the undissociated weak acid.
  • A larger \(Ka\) indicates a stronger weak acid because it dissociates more readily.
  • Weak acids with low \(Ka\) values have higher pKa and dissolve less in water, affecting their titration curve.
Understanding \(Ka\) helps predict the behavior of the acid in different chemical environments and is key in pH calculations during titrations.

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Most popular questions from this chapter

In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cd}^{2+}\), (b) \(\mathrm{Cr}(\mathrm{OH})_{2}\) and \(\mathrm{Fe}(\mathrm{OH})_{3}\), (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{K}^{4}\), (d) \(\mathrm{Ag}^{*}\) and \(\mathrm{Mn}^{2+}\), Suggest how each mixture might be separated.

From the value of \(K_{f}\) listed in Table \(17.1,\) calculate the concentration of \(\mathrm{Ni}^{2}(a q)\) and \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) that are present at equilibrium after dissolving 1.25 \(\mathrm{g} \mathrm{NiCl}_{2}\) in 100.0 \(\mathrm{mL}\) of 0.20 \(\mathrm{MN} \mathrm{H}_{3}(a q) .\)

(a) A 0.1044-g sample of an unknown monoprotic acid requires \(22.10 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{NaOH}\) to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the pH of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89\). What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid.

A solution contains three anions with the following concentrations: \(0.20 \mathrm{MCrO}_{4}^{2-}, 0.10 \mathrm{MCO}_{3}^{2-}\), and \(0.010 \mathrm{M} \mathrm{Cl}^{-}\). If a dilute \(\mathrm{AgNO}_{3}\) solution is slowly added to the solution, what is the first compound to precipitate: \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\left(K_{3 p}=1.2 \times 10^{-12}\right)\). \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\left(K_{\mathrm{p}}=8.1 \times 10^{-12}\right)\), or \(\mathrm{AgCl}\left(K_{\text {sp }}=1.8 \times 10^{-10}\right)\) ?

Which of the following solutions is a buffer? (a) \(0.10 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(0.10 \mathrm{MCH}_{3} \mathrm{COONa}\), (b) \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\), (c) \(0.10 \mathrm{M} \mathrm{HCl}\) and \(0.10 \mathrm{M} \mathrm{NaCl}\), (d) both a and \(c_{1}\) (e) all of a, b, and \(c\).

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