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Chapter 16: Problem 143

The titration of \(0.02500 \mathrm{~L}\) of a diprotic acid solution with \(0.1000\) M NaOH requires \(34.72 \mathrm{~mL}\) of titrant to reach the second equivalence point. The \(\mathrm{pH}\) is \(3.95\) at the first equivalence point and \(9.27\) at the second equivalence point. If the acid solution contained \(0.2015 \mathrm{~g}\) of the acid, what is the molar mass, \(\mathrm{p} K_{\mathrm{a} 1}\), and \(\mathrm{p} K_{\mathrm{a} 2}\) of the acid?

Short Answer

Expert verified
Molar mass 鈮� 116.05 g/mol, \(\mathrm{p}K_{\mathrm{a}1} = 3.95\), \(\mathrm{p}K_{\mathrm{a}2} = 9.27\).

Step by step solution

01

Find Moles of NaOH

Calculate the moles of NaOH used to reach the second equivalence point. Use the formula: \( ext{Moles of NaOH} = ext{Volume (L)} \times ext{Molarity (M)} \). First, convert \(34.72 \text{ mL}\) to liters: \(34.72 \text{ mL} = 0.03472 \text{ L}\). Now calculate: \( \text{Moles of NaOH} = 0.03472 \text{ L} \times 0.1000 \text{ M} = 0.003472 \text{ moles}\).
02

Moles of Acid

At the second equivalence point, the moles of NaOH used equals the moles of the diprotic acid because each mole of diprotic acid reacts with two moles of NaOH. Therefore, the moles of the diprotic acid is half of the moles of NaOH: \(0.003472 \div 2 = 0.001736 \text{ moles}\).
03

Calculate Molar Mass of the Acid

Use the formula: \( ext{Molar mass} = \frac{\text{Mass of acid (g)}}{\text{Moles of acid}} \). Insert the values: \( \text{Molar mass} = \frac{0.2015 \text{ g}}{0.001736 \text{ moles}} \approx 116.05 \text{ g/mol}\).
04

Evaluate \(\mathrm{p}K_{\mathrm{a}1}\) and \(\mathrm{p}K_{\mathrm{a}2}\)

For a diprotic acid, \(\mathrm{p}K_{\mathrm{a}1}\) can be estimated by the pH at the first equivalence point, and \(\mathrm{p}K_{\mathrm{a}2}\) by the pH at the second. This gives us \(\mathrm{p}K_{\mathrm{a}1} = 3.95\) and \(\mathrm{p}K_{\mathrm{a}2} = 9.27\).

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Key Concepts

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

Equivalence Point
In the world of titrations, the equivalence point is a critical concept. It marks the stage in a titration where the quantity of titrant added is exactly enough to completely neutralize the analyte solution. For a diprotic acid, there are two equivalence points because the acid releases two protons (H鈦� ions), requiring two separate reactions.
At each equivalence point, the amount of base (NaOH in this scenario) matches the quantity of dissociable protons from the acid. This is where the first indication occurs that the acidic solution has been neutralized once, and later twice, leading to two key pH values as the acid becomes completely neutralized. Understanding equivalence points is fundamental to accurately analyzing results from an acid-base titration.
Molar Mass Calculation
Molar mass provides a gateway to understanding the composition of substances at a molecular level. It is defined as the mass of one mole of a substance. Here, we calculated it by using the total mass of the acid divided by the moles of acid present. With the information provided in the exercise, the molar mass calculation involves:
  • Finding the moles of the acid through titration results.
  • Using the mass given for the sample of acid.
We determined that the molar mass of the acid is approximately 116.05 g/mol, identifying how many grams a single mole of this diprotic acid weighs. It's a crucial step to connect chemical processes at the macroscopic scale to reactions that happen at an atomic level.
pK_a Values
pK鈧� values reflect the strength of an acid, telling us how easily it can donate a proton. In this exercise, we learned about two pK鈧� values since the acid is diprotic. At each equivalence point, we can estimate the pK鈧� values from the pH readings.
  • pK鈧愨倎 at the first equivalence point was determined to be 3.95.
  • pK鈧愨倐 at the second equivalence point was set at 9.27.
These values indicate the acid's propensity to donate its protons. Lower pK鈧� values mean a stronger acid capable of releasing protons more easily, while higher values indicate a weaker proton donation capability. Understanding these values helps in predicting acid behavior in various chemical contexts.
Acid-Base Titration
Acid-base titration is a laboratory method used to determine the concentration of an acid or base in a solution. It involves gradually adding a titrant (a solution of known concentration) to a sample solution until the reaction reaches the equivalence point.
This method serves not only to find concentrations but also to calculate molar masses and understand the dissociation constants of acids.
  • The careful addition of NaOH to the diprotic acid allowed us to reach both equivalence points.
  • Observing these points provided data for calculating the molar mass and pK鈧� values.
Mastery of acid-base titration techniques is essential for any chemist, as it combines theoretical insights with practical skills to analyze and comprehend various chemical substances.

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Most popular questions from this chapter

Aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, K_{\mathrm{b}}=4.3 \times 10^{-10}\right)\) is a weak base used in the manufacture of dyes. Calculate the value of \(K_{\mathrm{n}}\) for the neutralization of aniline by vitamin \(\mathrm{C}\) (ascorbic acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}, K_{\mathrm{a}}=8.0 \times 10^{-5}\) ). Does much aniline remain at equilibrium?

A saturated solution of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) has \(\left[\mathrm{Ca}^{2+}\right]=2.01 \times 10^{-8} \mathrm{M}\) and \(\left[\mathrm{PO}_{4}^{3-}\right]=1.6 \times 10^{-5} \mathrm{M}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

The \(\mathrm{pH}\) of a solution of \(\mathrm{HN}_{3}\left(K_{\mathrm{a}}=1.9 \times 10^{-5}\right)\) and \(\mathrm{NaN}_{3}\) is \(4.86\). What is the molarity of \(\mathrm{NaN}_{3}\) if the molarity of \(\mathrm{HN}_{3}\) is \(0.016 \mathrm{M}\) ?

Consider the titration of \(50.0 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HA}\left(K_{\mathrm{a}}=\right.\) \(1.0 \times 10^{-4}\) ) with 0.010 M NaOH. (a) Sketch the pH titration curve, and label the equivalence point. (b) How many milliliters of \(0.010 \mathrm{M} \mathrm{NaOH}\) are required to reach the equivalence point? (c) Is the \(\mathrm{pH}\) at the equivalence point greater than, equal to, or less than \(7 ?\) (d) What is the pH exactly halfway to the equivalence point?

Assume that \(40.0 \mathrm{~mL}\) of \(0.0800 \mathrm{MH}_{2} \mathrm{SO}_{3}\left(K_{\mathrm{a} 1}=1.5 \times 10^{-2},\right.\), \(\left.K_{\mathrm{a} 2}=6.3 \times 10^{-8}\right)\) is titrated with \(0.160 \mathrm{M} \mathrm{NaOH}\). Calculate the \(\mathrm{pH}\) after addition of the following volumes of \(0.160 \mathrm{M} \mathrm{NaOH}\) : (a) \(20.0 \mathrm{~mL}\) (b) \(30.0 \mathrm{~mL}\) (c) \(35.0 \mathrm{~mL}\)
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