Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 40

Calculate the pH at the points in the titration of \(25.00 \mathrm{mL}\) of \(0.160 \mathrm{M} \mathrm{HCl}\) when (a) \(10.00 \mathrm{mL}\) and \((\mathrm{b}) 15.00 \mathrm{mL}\) of 0.242 M KOH have been added.

Short Answer

Expert verified
The pH at the points in the titration of \(0.275 \, \mathrm{M} \, \mathrm{KOH}\) when (a) \(15.00 \, \mathrm{mL}\) and (b) \(20.00 \, \mathrm{mL}\) of \(0.350 \, \mathrm{M} \, \mathrm{HCl}\) have been added are calculated based on the formulas outlined above.

Step by step solution

01

Calculate the initial moles of KOH

This value is given by the formula \(\text{Molarity (M)} \times \text{Volume (L)} = \text{Moles}\).In this case, \(0.275 \, \mathrm{M} \times 0.020 \, \mathrm{L} = 0.0055 \, \mathrm{moles}\) of KOH.
02

Calculate the moles of HCl added (Part a)

Again, we use the formula \(\text{Molarity (M)} \times \text{Volume (L)} = \text{Moles}\).For part (a), \(0.350 \, \mathrm{M} \times 0.015 \, \mathrm{L} = 0.00525 \, \mathrm{moles}\) of HCl.
03

Determine nature of solution and calculate resulting pH (Part a)

Subtract moles of HCl from moles of KOH to see if the solution is acidic or basic.In this case, \(0.0055 - 0.00525 = 0.00025 \, \mathrm{moles}\) of KOH are left; the solution is basic.To find the pH, we first find the pOH by dividing the remaining moles of KOH by the total volume of the solution (in this case, \(20.00 + 15.00 = 35.00 \, \mathrm{ml}\) or \(0.035 \, \mathrm{L}\), then use \(\text{pOH} = -\log [\text{OH}^-]\) to find the pOH, and finally use \(\text{pH} + \text{pOH} = 14\) to find the pH.
04

Calculate the moles of HCl added (Part b)

For part (b), we calculate moles of HCl as follows: \(0.350 \, \mathrm{M} \times 0.020 \, \mathrm{L} = 0.0070 \, \mathrm{moles}\) of HCl.
05

Determine nature of solution and calculate resulting pH (Part b)

In this case, after we subtract moles of KOH from moles of HCl (\(0.00700 - 0.0055 = 0.0015 \, \mathrm{moles}\)), we see that the solution is acidic as we have excess \(\mathrm{HCl}\). Calculate the moles of H^+ ions left by dividing moles of HCl left by the total volume of the solution, which is now \(20.00 + 20.00 = 40.00 \, \mathrm{ml}\) or \(0.040 \, \mathrm{L}\). Use \(\text{pH} = -\log [\text{H}^+]\) to find the pH.

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

The \(\mathrm{pH}\) of ocean water depends on the amount of atmospheric carbon dioxide. The dissolution of carbon dioxide in ocean water can be approximated by the following chemical reactions (Henry's Law constant for \(\left.\mathrm{CO}_{2} \text { is } K_{\mathrm{H}}=\left[\mathrm{CO}_{2}(\mathrm{aq})\right] /\left[\mathrm{CO}_{2}(\mathrm{g})\right]=0.8317 .\right)\) \(\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{aq})\) \(\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Ca}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{-}(\mathrm{aq})\) \(\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{CO}_{3}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{HCO}_{3}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) \(\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{HCO}_{3}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1)\) (a) Use the equations above to determine the hydronium ion concentration as a function of \(\left[\mathrm{CO}_{2}(\mathrm{g})\right]\) and \(\left[\mathrm{Ca}^{2+}\right]\) (b) During preindustrial conditions, we will assume that the equilibrium concentration of \(\left[\mathrm{CO}_{2}(\mathrm{g})\right]=280\) ppm and \(\left[\mathrm{Ca}^{2+}\right]=10.24 \mathrm{mM} .\) Calculate the \(\mathrm{pH}\) of a sample of ocean water.

Rather than calculate the \(\mathrm{pH}\) for different volumes of titrant, a titration curve can be established by calculating the volume of titrant required to reach certain \(\mathrm{pH}\) values. Determine the volumes of \(0.100 \mathrm{M} \mathrm{NaOH}\) required to reach the following pH values in the titration of \(20.00 \mathrm{mL}\) of \(0.150 \mathrm{M} \mathrm{HCl}: \mathrm{pH}=\) (a) 2.00 (b) \(3.50 ;\) (c) \(5.00 ;\) (d) \(10.50 ;\) (e) \(12.00 .\) Then plot the titration curve.

Calculate the \(\mathrm{pH}\) of a \(0.5 \mathrm{M}\) solution of \(\mathrm{Ca}(\mathrm{HSe})_{2}\), given that \(\mathrm{H}_{2}\) Se has \(K_{\mathrm{a}_{1}}=1.3 \times 10^{-4}\) and \(K_{\mathrm{a}_{2}}=1 \times 10^{-11}\)

In the use of acid-base indicators, (a) Why is it generally sufficient to use a single indicator in an acid-base titration, but often necessary to use several indicators to establish the approximate pH of a solution? (b) Why must the quantity of indicator used in a titration be kept as small as possible?

Explain whether the equivalence point of each of the following titrations should be below, above, or at pH 7: (a) \(\mathrm{NaHCO}_{3}(\text { aq) titrated with } \mathrm{NaOH}(\mathrm{aq}) ; \text { (b) } \mathrm{HCl}(\mathrm{aq})\) titrated with \(\mathrm{NH}_{3}(\mathrm{aq}) ;\) (c) KOH(aq) titrated with HI(aq).
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic 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.

Sign-up for free