/*! 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 85 Calculate the \(\mathrm{pH}\) of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 85

Calculate the \(\mathrm{pH}\) of a solution obtained by mixing \(500.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{NH}_{3}\) with \(200.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{HCl}\).

Short Answer

Expert verified
The pH of the solution is approximately 9.08.

Step by step solution

01

Determine Moles of Reactants

First, calculate the number of moles of each reactant in the solution. For \(\text{NH}_3\), the moles are calculated as follows: \[moles \, of \, NH_3 = 0.10 \, M \times 0.500 \, L = 0.050 \, moles \]For \(\text{HCl}\), the moles are calculated as follows:\[moles \, of \, HCl = 0.15 \, M \times 0.200 \, L = 0.030 \, moles\]
02

Determine Limiting Reactant

Identify the limiting reactant in the solution by comparing the mole ratio. Since HCl is in lesser moles (\(0.030\)), it will completely react with part of the NH鈧�, which has \(0.050\) moles.
03

Calculate Excess NH鈧� Moles

Subtract moles of HCl from moles of NH鈧� to find the moles of NH鈧� remaining after the reaction: \[\text{Moles of } NH_3 \, remaining = 0.050 - 0.030 = 0.020 \, moles\]
04

Total Volume of Final Solution

Calculate the total volume of the resulting solution by adding the volumes of the two solutions:\[V_{\text{total}} = 500.0 \, mL + 200.0 \, mL = 700.0 \, mL = 0.700 \, L\]
05

Calculate Concentration of NH鈧�

Determine the concentration of NH鈧� in the final solution using the excess moles and total volume:\[[NH_3] = \frac{0.020 \, moles}{0.700 \, L} = 0.02857 \, M\]
06

Calculate pH of the Solution

Since NH鈧� is a weak base, calculate the pH using the base dissociation constant \(K_b = 1.8 \times 10^{-5}\) for NH鈧�. First, find \([OH^-]\):\[[NH_4^+]_{\text{formed}} = 0.030 \, moles / 0.700 \, L = 0.0429 \, M\]\[K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = 1.8 \times 10^{-5}\]Assuming \([OH^-] = x\),\[1.8 \times 10^{-5} = \frac{0.0429 \times x}{0.02857}\]Solving for \(x\) gives \(x \, \approx 1.2 \times 10^{-5} \, M\).Finally, find \(\text{pOH}\, \text{and}\, \text{pH}\):\[pOH = - \log(1.2 \times 10^{-5}) \approx 4.92\]\[pH = 14 - 4.92 = 9.08\]
07

Validate Assumptions

Check that our approximation for \([OH^-]\) is valid by comparing \(K_b\) with initial concentration of \(NH_3\). Our assumptions hold since the calculated hydroxide concentration is relatively low.

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.

Acid-Base Reaction
In the realm of chemistry, an acid-base reaction occurs when an acid donates a proton (H鈦�) to a base. In the provided exercise, we see such a reaction between the base ammonia (\( \mathrm{NH}_3\)) and the acid hydrochloric acid (\( \mathrm{HCl}\)). When they mix, \( \mathrm{HCl}\) releases hydrogen ions, which get accepted by the ammonia, converting it to ammonium ions (\( \mathrm{NH}_4^+\)). This fundamental concept of proton transfer underpins the reaction taking place.
Key points to remember:
  • An acid releases protons, like \( \mathrm{HCl}\), which turns into chloride ions after losing a proton.
  • A base accepts protons, like \( \mathrm{NH}_3\), which becomes \( \mathrm{NH}_4^+\) upon gaining a proton.
Understanding this proton exchange is crucial to grasp the mechanics of the given acid-base reaction.
Limiting Reactant
The concept of a limiting reactant in a chemical reaction tells us which reactant will be consumed first, effectively stopping the reaction from proceeding further. Identifying the limiting reactant is pivotal because it determines the maximum amount of product that can be formed.
In our exercise, \( \mathrm{HCl}\) is the limiting reactant. Here鈥檚 why:
  • There are \( 0.030 \, ext{moles} \) of \( \mathrm{HCl}\) compared to \( 0.050 \, ext{moles} \) of \( \mathrm{NH}_3\).
  • \( \mathrm{HCl}\) will react completely with a portion of \( \mathrm{NH}_3\), limiting the amount of the reaction.
Once the \( \mathrm{HCl}\) is used up, the reaction will halt, and any excess \( \mathrm{NH}_3\) remains unreacted. Grasping this concept allows us to predict how reactions will proceed and what amounts of products or leftover reactants we might expect.
Base Dissociation Constant
The base dissociation constant, \( K_b\), quantifies the strength of a base in solution. It shows how well a base can accept protons by expressing the equilibrium between the base and its conjugate acid in water. The larger the \( K_b\), the stronger the base.In this exercise, \( \mathrm{NH}_3\), a weak base, has a \( K_b\) value of \( 1.8 \times 10^{-5} \). This small \( K_b\) indicates that ammonia does not dissociate much in water, forming relatively few hydroxide ions:
  • The equilibrium involves the formation of \( \mathrm{NH}_4^+\) and \( \mathrm{OH^-}\).
  • The expression \( K_b = \frac{[\mathrm{NH}_4^+][\mathrm{OH^-}]}{[\mathrm{NH}_3]} \) helps calculate the solution's hydroxide concentration.
It's crucial to understand \( K_b\) as it helps predict the resulting \( \mathrm{pH}\), reshaping how we see ammonia's behavior in our mixed solution.
Solution Concentration
Solution concentration describes the amount of solute present in a certain volume of solvent, often expressed in molarity (M). It helps us determine reactant and product quantities throughout chemical reactions.
In our scenario, the concentration of \( \mathrm{NH}_3\) changes post-reaction. Initially in the solution, we had specific concentrations of \( \mathrm{NH}_3\) and \( \mathrm{HCl}\):
  • Initial \( \mathrm{NH}_3\) was \( 0.10 \, ext{M} \).
  • Initial \( \mathrm{HCl}\) was \( 0.15 \, ext{M} \).
  • After reaction, excess \( \mathrm{NH}_3\) left led to a new concentration of \( 0.02857 \, ext{M} \).
Adjusting concentrations based on reaction completion allows us to further calculate properties of the solution, such as its \( \mathrm{pH}\). Monitoring these changes is essential for accurate application of acid-base chemistry principles in calculations.

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

Ionization of the first proton from \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is complete \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right.\) is a strong acid); the acid-ionization constant for the second proton is \(1.1 \times 10^{-2}\). (a) What would be the approximate hydronium-ion concentration in \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) if ionization of the second proton were ignored? (b) The ionization of the second proton must be considered for a more exact answer, however. Calculate the hydronium-ion concentration in \(0.100\) \(M \mathrm{H}_{2} \mathrm{SO}_{4}\), accounting for the ionization of both protons.

What is the \(\mathrm{pH}\) of a solution in which \(40 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{NaOH}\) is added to \(25 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HCl} ?\)

Malic acid is a weak diprotic organic acid with \(K_{a 1}=\) \(4.0 \times 10^{-4}\) and \(K_{a 2}=9.0 \times 10^{-6}\) a. Letting the symbol \(\mathrm{H}_{2} \mathrm{~A}\) represent malic acid, write the chemical equations that represent \(K_{a 1}\) and \(K_{a 2} .\) Write the chemical equation that represents \(K_{a 1} \times K_{a 2}\) b. Qualitatively describe the relative concentrations of \(\mathrm{H}_{2} \mathrm{~A}\), \(\mathrm{HA}^{-}, \mathrm{A}^{2-}\), and \(\mathrm{H}_{3} \mathrm{O}^{+}\) in a solution that is about one molar in malic acid. c. Calculate the \(\mathrm{pH}\) of a \(0.0100 \mathrm{M}\) malic acid solution and the equilibrium concentration of \(\left[\mathrm{H}_{2} \mathrm{~A}\right]\). d. What is the \(A^{2-}\) concentration?

A chemist prepares dilute solutions of equal molar concentrations of \(\mathrm{NH}_{3}, \mathrm{NH}_{4} \mathrm{Br}, \mathrm{NaF}\), and \(\mathrm{NaCl}\). Rank these solutions from highest \(\mathrm{pH}\) to lowest \(\mathrm{pH}\).

A \(0.288-\mathrm{g}\) sample of an unknown monoprotic organic acid is dissolved in water and titrated with a \(0.115 \mathrm{M}\) sodium hydroxide solution. After the addition of \(17.54 \mathrm{~mL}\) of base, a \(\mathrm{pH}\) of \(4.92\) is recorded. The equivalence point is reached when a total of \(33.83 \mathrm{~mL}\) of \(\mathrm{NaOH}\) is added. a. What is the molar mass of the organic acid? b. What is the \(K_{a}\) value for the acid? The \(K_{a}\) value could have been determined very easily if a pH measurement had been made after the addition of \(16.92 \mathrm{~mL}\) of \(\mathrm{NaOH}\). Why?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

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.