/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Consider the titration of 100.0 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 19

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{M} \mathrm{H}_{3} \mathrm{AsO}_{4}\) by 0.10 \(M \mathrm{NaOH}\) . What are the major species present at 50.0 \(\mathrm{mL}\) of \(\mathrm{NaOH}\) added? How would you calculate the pH at this point? Answer the same questions for 150.0 \(\mathrm{mL}\) of NaOH added. At what volume of NaOH added does pH \(=\mathrm{p} K_{\mathrm{a}_{\mathrm{i}}} ?\)

Short Answer

Expert verified
At 50.0 mL of NaOH added, the major species present are H鈧傾sO鈧� and H鈧侫sO鈧勨伝. To calculate the pH at this point, we use the first acidity constant (K鈧愨倎) given and the equilibrium concentrations of H鈧傾sO鈧� and H鈧侫sO鈧勨伝. At 150.0 mL of NaOH added, the major species present are H鈧傾sO鈧�, H鈧侫sO鈧勨伝, and HAsO鈧劼测伝. To calculate the pH at this point, we use the second acidity constant (K鈧愨倐) and the respective concentrations of the species. To find the volume of NaOH added, where pH = pK鈧愥耽, we must determine the half-equivalence points, where [A鈦籡 = [HA], for each dissociation reaction and compare the pH and pK鈧� values at these points.

Step by step solution

01

Analyzing the dissociation of H鈧傾sO鈧� and corresponding reactions with NaOH

Since H鈧傾sO鈧� is a triprotic acid, we need to consider its dissociation and the corresponding reactions with NaOH, which acts as a strong base. The dissociation reactions are as follows: 1) H鈧傾sO鈧� + OH鈦� 鈫� H鈧侫sO鈧勨伝 + H鈧侽 2) H鈧侫sO鈧勨伝 + OH鈦� 鈫� HAsO鈧劼测伝 + H鈧侽 3) HAsO鈧劼测伝 + OH鈦� 鈫� AsO鈧劼斥伝 + H鈧侽
02

Calculate the moles of the species present at different volumes of NaOH added.

First, let's determine the initial number of moles of H鈧傾sO鈧� and NaOH at each given volume. Since the initial concentration of H鈧傾sO鈧� (as well as NaOH) is given in M (molarity), we can multiply it by the volume in liters to obtain the number of moles. Moles of H鈧傾sO鈧� = 0.10 M 脳 100.0 mL 脳 (1 L/1000 mL) = 0.01 mol For 50.0 mL of added NaOH: Moles of NaOH = 0.10 M 脳 50.0 mL 脳 (1 L/1000 mL) = 0.005 mol For 150.0 mL of added NaOH: Moles of NaOH = 0.10 M 脳 150.0 mL 脳 (1 L/1000 mL) = 0.015 mol
03

Determine the dominant reaction for each volume of NaOH added.

For 50.0 mL of NaOH: There is an excess of H鈧傾sO鈧�, so we can assume that the stoichiometry is 1:1 for the first reaction: H鈧傾sO鈧� + OH鈦� 鈫� H鈧侫sO鈧勨伝 + H鈧侽. For 150.0 mL of NaOH: There is an excess of NaOH, so after the first two moles of OH鈦� react with H鈧傾sO鈧� and form H鈧侫sO鈧勨伝, the remaining NaOH will react with H鈧侫sO鈧勨伝.
04

Calculate the pH at each volume of NaOH added using the dominant reactions and pK鈧� values.

At 50.0 mL of added NaOH: Since the first dissociation is dominant (1:1 stoichiometry), the solution contains the conjugate base H鈧侫sO鈧勨伝 with H鈧傾sO鈧�. At this point, we have to use the first acidity constant (K鈧愨倎): K鈧愨倎 = \([H^+][H鈧侫sO鈧勨伝]\)/\([H鈧傾sO鈧刔\) With given K鈧� values, we can find the pH value. For 150.0 mL of added NaOH: Here, both the first and second reactions are dominant. At this point, we have the conjugate base HAsO鈧劼测伝 along with H鈧傾sO鈧� and H鈧侫sO鈧勨伝. We use the second acidity constant (K鈧愨倐) to calculate the pH: K鈧愨倐 = \([H^+][HAsO鈧劼测伝]\)/\([H鈧侫sO鈧勨伝]\)
05

Find the volume of NaOH added for which pH is equal to pK鈧愥耽

As we know that pH = pK鈧� at the half-equivalence point, where \([A^-]\) = \([HA]\). For each dissociation reaction, we will find the volume of NaOH required for the half-equivalence point and then compare the pK鈧� value to the pH. If pH = pK鈧�, then we have the required volume at that point. Repeat the corresponding steps, considering the other dissociation steps (the second and third dissociation reactions). Hence, by following these steps, we can find the major species present, calculate the pH at different volumes of added NaOH, and find the volume of NaOH, where the pH equals the pK鈧愥耽 value.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Triprotic Acid
A triprotic acid is a type of polyprotic acid that can donate three protons (hydrogen ions, \(H^+\)) per molecule. This ability to donate three protons results in three sequential dissociation reactions, each with its own equilibrium constant, referred to as \(K_a\).

In the case of the exercise, arsenic acid (\(H_3AsO_4\)) is the triprotic acid that undergoes the following dissociation steps:
  • \(H_3AsO_4 + OH^- \rightarrow H_2AsO_4^- + H_2O\) (first dissociation)
  • \(H_2AsO_4^- + OH^- \rightarrow HAsO_4^{2-} + H_2O\) (second dissociation)
  • \(HAsO_4^{2-} + OH^- \rightarrow AsO_4^{3-} + H_2O\) (third dissociation)
Each step progresses when a base like NaOH is added, providing hydroxide ions (\(OH^-\)) to reactors. Understanding these dissociation steps is crucial to predicting the behavior of a triprotic acid in a titration and its resulting pH changes.
pH Calculation
Calculating the pH during the titration of a triprotic acid involves understanding the concentration of protons and different ionic species present in solution. During each dissociation step, the dominant species will change, which directly affects the pH.

To calculate pH:
  • Identify the major species present after each addition of NaOH, based on stoichiometry and dissociation steps known for triprotic acids.
  • Use the relevant equilibrium constant \(K_a\) for the specific dissociation step that is dominant at a particular point in the titration.
  • Apply the formula: \(K_a = \frac{[H^+][A^-]}{[HA]}\) to solve for \([H^+]\), then use \(pH = -\log[H^+]\) to find the pH.
For example, at the first half-equivalence point, you primarily consider the first dissociation. As more OH鈦� is added, subsequent dissociation reactions become significant, and thus, different \(K_a\) values must be used for precise pH calculations.
Half-Equivalence Point
The half-equivalence point in a titration of a polyprotic acid is where half of the acid has been neutralized. At this point, the concentration of the acid equals the concentration of its conjugate base, making it significant for pH calculation.

Key points to consider:
  • At the half-equivalence point, \(pH = pK_a\) for that particular dissociation. This is an important property as it allows easy calculation and interpretation of the titration curve.
  • For each dissociation step of a triprotic acid, there is a corresponding half-equivalence point, each signifying a stage in the titration where the triprotic acid partially neutralizes.
  • To find it experimentally, you need to determine at what volume of \(NaOH\) the pH equals the \(pK_a\), assisting in recognizing the buffer region in a titration curve.
Understanding half-equivalence points aids in evaluating buffer capacities in solutions and designing buffer systems.
Dissociation Reactions
Dissociation reactions are the processes by which an acid releases its protons to form its conjugate base in a solution. For a triprotic acid, like \(H_3AsO_4\), these reactions occur in a stepwise manner, each step possessing unique characteristics.
  • Each dissociation is characterized by its own equilibrium constant represent by \(K_a\). This determines the strength of each step of dissociation.
  • Successful predictions of acid-base reactions rely on understanding each specific dissociation equilibrium, noting how each subsequent step requires progressively less energy compared to the first.
During a titration:
  • The reaction proceeds through each dissociation phase sequentially as a base is added.
  • This helps in determining not just the resulting species but also helps ascertain the pH during different phases of titration.
  • Practically observing these reactions can be seen by plotting a titration curve, which displays the characteristic inflection points associated with each dissociation transition.
Understanding these mechanisms is essential for accurately modeling the behavior of triprotic systems in chemical equilibria.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Th pH of blood is steady at a value of approximately 7.4 as a result of the following equilibrium reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \leftrightharpoons \mathrm{HCO}_{3}-(a q)+\mathrm{H}^{+}(a q) $$ The actual buffer system in blood is made up of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) and \(\mathrm{HCO}_{3}\) - One way the body keeps the pH of blood at 7.4 is by regulating breathing. Under what blood ph conditions will the body increase breathing and under what blood pH conditions will the body decrease breathing? Explain.

A \(0.400-M\) solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur?

Consider the titration of 50.0 \(\mathrm{mL}\) of 0.10\(M \mathrm{H}_{3} \mathrm{A}\left(K_{\mathrm{a}},=\right.\) \(5.0 \times 10^{-4}, K_{\mathrm{a}_{2}}=1.0 \times 10^{-8}, K_{\mathrm{a}_{2}}=1.0 \times 10^{-11}\) ) titrated by 0.10\(M \mathrm{KOH}\) a. Calculate the pH of the resulting solution at 125 \(\mathrm{mL}\) of KOH added. b. At what volume of KOH added does pH \(=3.30 ?\) c. At 75.0 \(\mathrm{mL}\) of KOH added, is the solution acidic or basic?

Another way to treat data from a pH titration is to graph the absolute value of the change in \(\mathrm{pH}\) per change in milliliters added versus milliliters added (\DeltapH/ \(\Delta \mathrm{mL}\) versus \(\mathrm{mL}\) added). Make this graph using your results from Exercise \(67 .\) What advantage might this method have over the traditional method for treating titration data?

Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) with 0.100 \(\mathrm{M}\) \(\mathrm{HCl} .\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.