/*! 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 47 In the titration of \(20.00 \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 47

In the titration of \(20.00 \mathrm{mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH},\) calculate the number of milliliters of \(0.200 \mathrm{M} \mathrm{HCl}\) that must be added to reach a pH of (a) \(12.55,\) (b) \(10.80,\) (c) 4.25

Short Answer

Expert verified
The volumes of 0.200 M HCl solution required to reach the pH values of 12.55, 10.80, and 4.25 are respectively calculated in steps 2 through 4. The calculations need to be repeated separately for each of the given pH values.

Step by step solution

01

Understanding the reagent relationship

The reaction that occurs during the titration of NaOH with HCl is a neutralization reaction, which produces water and sodium chloride. This reaction can be represented as \( \mathrm{NaOH} + \mathrm{HCl} \rightarrow \mathrm{H_2 O} + \mathrm{NaCl} \). From this reaction, it can be observed that NaOH and HCl react in a 1:1 ratio. Hence, the concentration and volume of NaOH can be used to calculate the moles of NaOH.
02

Calculating the moles of NaOH

The moles of NaOH can be calculated using the formula: moles = concentration * volume. The volume of NaOH provided is 20.00 mL, which needs to be converted to liters; hence, the volume is 0.02 L. Multiplying the volume (0.02 L) and the concentration (0.175 M), we get 0.0035 moles of NaOH.
03

Calculating the moles of HCl for respective pH values

The pH at which the volume required is to be calculated is given, so we can use the pH formula to find the concentration of HCl at that pH. The formula for the pH is \( \mathrm{pH} = -\log (\mathrm{[H+]}) \). So we can rearrange this to get the hydrogen ion concentration [H+] = \(10^{-pH}\). Calculate [H+] for each of the given pH values.
04

Calculating the volume of HCl required

After the [H+] concentration is known, subtract the concentration of HCl required before the equivalence point from the remaining moles of OH- ions left after the equivalence point. Then, divide these moles by the concentration of the HCl solution to find the volume. Convert the volume to mL by multiplying by 1000.
05

Repeat calculations for given pH levels

Repeat steps 3 and 4 to calculate the volume of HCl required to reach each of the other two given pH levels.

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.

Neutralization Reaction
When performing a titration between an acid and a base, you are witnessing a neutralization reaction. In this process, an acid reacts with a base to produce water and a salt. In our case, sodium hydroxide (NaOH), a strong base, reacts with hydrochloric acid (HCl), a strong acid:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{H}_2\text{O} + \text{NaCl} \]
  • NaOH and HCl are mixed in a 1:1 molar ratio, meaning they neutralize each other perfectly in equivalent amounts.
  • The balanced equation tells us that one mole of NaOH neutralizes one mole of HCl, resulting in the formation of water (H鈧侽) and salt (NaCl).
This reaction is crucial in titration experiments as it helps determine an unknown concentration by adding a known concentration of another reactant until the reaction is complete.
pH Calculation
The pH scale measures how acidic or basic a solution is, with values ranging from 0 to 14. A pH of 7 is neutral, below 7 is acidic, and above 7 is basic. To calculate pH during titration, understanding the relationship between the concentration of hydrogen ions \(\text{[H鈦篯}\) and pH is important.

The formula used is:
\[ \text{pH} = -\log(\text{[H鈦篯}) \]
  • Rearranging the formula gives us \(\text{[H鈦篯} = 10^{-\text{pH}}\) to find the concentration of hydrogen ions from a given pH.
  • For example, if the pH is 4.25, \(\text{[H鈦篯}\) would be \(10^{-4.25}\).
During titration, these calculations determine how much acid or base to add to achieve the desired pH level, analyzing the concentration before and after reaching the equivalence point.
Molarity
Molarity is the concentration of a solution, expressed as the number of moles of solute per liter of solution. It's a key player in titration calculations.

Understanding molarity allows you to relate volumes and concentrations to the stoichiometry of the reaction.
  • Molarity (M) = \( \frac{\text{moles of solute}}{\text{liter of solution}} \)
  • For instance, in the given problem, 0.175 M NaOH means there are 0.175 moles of NaOH in 1 liter of solution.
  • To find moles, multiply molarity by the volume of the solution in liters. So, \( 0.175 \text{ M} \times 0.02 \text{ L} = 0.0035 \text{ moles} \).
This concept ties into the neutralization where the moles of acids and bases react to reach equivalence, and helps calculate unknown concentrations by knowing volumes and the molar ratio.

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

Thymol blue indicator has \(t w o\) pH ranges. It changes color from red to yellow in the pH range from 1.2 to 2.8, and from yellow to blue in the pH range from 8.0 to 9.6. What is the color of the indicator in each of the following situations? (a) The indicator is placed in \(350.0 \mathrm{mL}\) of \(0.205 \mathrm{M} \mathrm{HCl}\) (b) To the solution in part (a) is added \(250.0 \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{NaNO}_{2}\) (c) To the solution in part (b) is added \(150.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) (d) To the solution in part (c) is added \(5.00 \mathrm{g} \mathrm{Ba}(\mathrm{OH})_{2}\)

To repress the ionization of formic acid, HCOOH(aq), which of the following should be added to the solution? (a) \(\mathrm{NaCl} ;\) (b) \(\mathrm{NaOH}\); (c) \(\mathrm{NaHCOO}\); (d) \(\mathrm{NaNO}_{3}\)

The neutralization of \(\mathrm{NaOH}\) by \(\mathrm{HCl}\) is represented in equation (1), and the neutralization of \(\mathrm{NH}_{3}\) by HCl in equation (2). 1. \(\mathrm{OH}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O} \quad K=?\) 2\. \(\mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \quad K=?\) (a) Determine the equilibrium constant \(K\) for each reaction. (b) Explain why each neutralization reaction can be considered to go to completion.

To increase the ionization of formic acid, \(\mathrm{HCOOH}(\mathrm{aq})\) which of the following should be added to the solution? (a) \(\mathrm{NaCl} ;\) (b) \(\mathrm{NaHCOO} ;(\mathrm{c}) \mathrm{H}_{2} \mathrm{SO}_{4} ;\) (d) \(\mathrm{NaHCO}_{3}\)

A solution is prepared that is \(0.150 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(0.250 \mathrm{M} \mathrm{NaHCOO}\) (a) Show that this is a buffer solution. (b) Calculate the pH of this buffer solution. (c) What is the final pH if 1.00 L of 0.100 M HCl is added to \(1.00 \mathrm{L}\) of this buffer solution?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

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.