Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 87

A \(0.4324 \mathrm{g}\) sample of a potassium hydroxide-lithium hydroxide mixture requires \(28.28 \mathrm{mL}\) of \(0.3520 \mathrm{M} \mathrm{HCl}\) for its titration to the equivalence point. What is the mass percent lithium hydroxide in this mixture?

Short Answer

Expert verified
0 %

Step by step solution

01

Calculate moles of HCl

Molarity is defined as the number of moles of solute per litre of solution. So, the amount of moles of HCl can be calculated by multiplying the molarity by the volume (in litres). Moles of \(HCl = 0.3520 \, M \times 0.02828 \, L = 0.00995456 \, mol\).
02

Determine moles of OH-

In a titration to the equivalence point with HCl, the moles of hydronium ion (H+) from HCl will equal the moles of hydroxide ion (OH-) from the base there was originally in the mixture. Hence, moles of OH- = 0.00995456 mol.
03

Calculate mass of each hydroxide

Potassium hydroxide (KOH) and lithium hydroxide (LiOH) both provide one mole of OH- per mole. We can now write an equation representing the total OH- from KOH and LiOH as: Mass = moles OH- \times molar mass = 0.00995456 mol \times 56.1 g/mol (for KOH) + 0.00995456 mol \times 23.95 g/mol (for LiOH) = 0.5586 g + 0.23833 g = 0.79693 g. This is the maximum total mass of KOH and LiOH that could be in the sample.
04

Calculate mass percent of LiOH

Knowing maximum total mass of KOH and LiOH that could be in the sample and the actual mass of the sample, we can now subtract to get the mass of the unknown component. Subtracting from the total mass yields an estimate of the mass of LiOH: Mass of LiOH = total mass of sample - mass of KOH = 0.4324 g - 0.5586 g = -0.1262 g. Since the calculated mass of LiOH is negative, this suggests that the sample doesn't contain LiOH. Therefore, the percent by mass of LiOH in the mixture would be 0.

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 number of moles of hydroxide ion in 0.300 L of \(0.0050 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is (a) \(0.0015 ;(\mathrm{b}) 0.0030 ;(\mathrm{c}) 0.0050\) (d) 0.010.

The exact neutralization of \(10.00 \mathrm{mL}\) of \(0.1012 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\text { aq })\) requires \(23.31 \mathrm{mL}\) of \(\mathrm{NaOH}\). What must be the molarity of the \(\mathrm{NaOH}(\mathrm{aq}) ?\) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1)\)

Which of the following aqueous solutions has the highest concentration of \(\mathrm{K}^{+}\) ? (a) \(0.0850 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4};\) (b) a solution containing \(1.25 \mathrm{g} \mathrm{KBr} / 100 \mathrm{mL} ;\) (c) a solution having \(8.1 \mathrm{mg} \mathrm{K}^{+} / \mathrm{mL}\).

Complete each of the following as a net ionic equation, indicating whether a precipitate forms. If no reaction occurs, so state. (a) $\mathrm{Na}^{+}+\mathrm{Br}^{-}+\mathrm{Pb}^{2+}+2 \mathrm{NO}_{3}^{-} \longrightarrow$ (b) $\mathrm{Mg}^{2+}+2 \mathrm{Cl}^{-}+\mathrm{Cu}^{2+}+\mathrm{SO}_{4}^{2-} \longrightarrow$ (c) $\mathrm{Fe}^{3+}+3 \mathrm{NO}_{3}^{-}+\mathrm{Na}^{+}+\mathrm{OH}^{-} \longrightarrow$

Briefly describe (a) half-equation method of balancing redox equations; (b) disproportionation reaction; (c) titration; (d) standardization of a solution.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

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