Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 64

A volume of \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) is titrated against a \(0.100 \mathrm{M} \mathrm{NH}_{3}\) solution added to it from a buret. Calculate the \(\mathrm{pH}\) values of the solution (a) after \(10.0 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) solution have been added, \((\mathrm{b})\) after \(25.0 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) solution have been added, (c) after \(35.0 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) solution have been added.

Short Answer

Expert verified
a) pH=1.2 , b) pH=9.25 , c) pH= 9.7. These values might slightly vary based on the exact pKa value used.

Step by step solution

01

Calculating the moles of HCl and NH3

First, calculate the number of moles of HCl in the 25.0 mL solution using the molarity formula M = n/V. Thus, n(HCl) = M x V = 0.100 mol/L x 0.025 L = 0.0025 mol. Similarly, the number of moles of NH3 added at each point can be calculated as n(NH3) = M x V.
02

Calculating the moles of NH3 for each given volume and determining the resulting species

Next, calculate the amount of NH3 in mol that was added in each case (10.0 mL, 25.0 mL, and 35.0 mL). In each case, deduct the mol of HCl present initially from moles of NH3 added. If the moles of NH3 are greater than HCl, then they will neutralize HCl and NH4Cl will form. If the moles of NH3 are less than HCl, then it will be a case of a weak base titrated with a strong acid.
03

Calculating the pH using the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation (pH = pKa + log([A-]/[HA])) can be applied, keeping in mind that A- and HA will be different for each part of the scenario. [A-] is the concentration for the excess substance after reaction while [HA] is the concentration for the substance which was initially higher and has been partially neutralized. pKa in case of NH3 is the pKa of the conjugate acid (NH4+) which is 9.25.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A quantity of \(0.560 \mathrm{~g}\) of \(\mathrm{KOH}\) is added to \(25.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). Excess \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is then added to the solution. What mass (in grams) of \(\mathrm{CO}_{2}\) is formed?

If \(\mathrm{NaOH}\) is added to \(0.010 \mathrm{MAl}^{3+}\), which will be the predominant species at equilibrium: \(\mathrm{Al}(\mathrm{OH})_{3}\) or \(\mathrm{Al}(\mathrm{OH})_{4}^{-} ?\) The \(\mathrm{pH}\) of the solution is \(14.00 .\left[K_{\mathrm{f}}\right.\) for \(\mathrm{Al}(\mathrm{OH})_{4}^{-}=2.0 \times 10^{33}\)

Which of these ionic compounds will be more soluble in acid solution than in water: (a) \(\mathrm{BaSO}_{4}\) (b) \(\mathrm{PbCl}_{2},\) (c) \(\mathrm{Fe}(\mathrm{OH})_{3}\) (d) \(\mathrm{CaCO}_{3}\) ? Explain.

In a group 1 analysis, a student obtained a precipitate containing both \(\mathrm{AgCl}\) and \(\mathrm{PbCl}_{2}\). Suggest one reagent that would allow her to separate \(\operatorname{AgCl}(s)\) from \(\mathrm{PbCl}_{2}(s)\)

Specify which of these systems can be classified as a buffer system: (a) \(\mathrm{KNO}_{2} / \mathrm{HNO}_{2}\) (b) \(\mathrm{KHSO}_{4} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (c) HCOOK/HCOOH.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

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.

Sign-up for free