Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 46

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.182 \mathrm{M} \mathrm{KOH},\) (b) \(3.165 \mathrm{~g}\) of \(\mathrm{KOH}\) in \(500.0 \mathrm{~mL}\) of solution, (c) \(10.0 \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{~mL}\) (d) a solution formed by mixing \(20.0 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}\).

Short Answer

Expert verified
(a) [OH⁻] = 0.182 M; pH = 13.26 (b) [OH⁻] = 0.1128 M; pH = 13.053 (c) [OH⁻] = 4.2 × 10⁻⁴ M; pH = 10.62 (d) [OH⁻] = 1.5467 × 10⁻² M; pH = 12.19

Step by step solution

01

Calculate the moles of OH⁻ ions

KOH is a strong base that dissociates completely in water according to the equation: KOH → K⁺ + OH⁻. Therefore, 1 mole of KOH produces 1 mole of OH⁻. The concentration of OH⁻ ions is the same as that of KOH: 0.182 M.
02

Calculate pH

Now we'll use the OH⁻ concentration to calculate the pH using the formula: pOH = -log[OH⁻] and pH = 14 - pOH. pOH = -log(0.182) = 0.74 pH = 14 - 0.74 = 13.26 (a) [OH⁻] = 0.182 M; pH = 13.26 (b) 3.165 g of KOH in 500.0 mL of solution
03

Calculate the moles of OH⁻ ions

First, we'll calculate the moles of KOH by dividing the mass by the molar mass of KOH (56.11 g/mol). moles of KOH = 3.165 g / 56.11 g/mol = 0.0564 mol Since 1 mole KOH produces 1 mole OH⁻, moles of OH⁻ = moles of KOH = 0.0564 mol
04

Calculate the concentration of OH⁻ ions

To find the concentration of OH⁻ ions, divide the moles of OH⁻ by the volume of the solution in liters. [OH⁻] = 0.0564 mol / 0.5 L = 0.1128 M
05

Calculate pH

pOH = -log(0.1128) = 0.947 pH = 14 - 0.947 = 13.053 (b) [OH⁻] = 0.1128 M; pH = 13.053 (c) 10.0 mL of 0.0105 M Ca(OH)₂ diluted to 500.0 mL
06

Calculate the moles of OH⁻ ions

The dissociation of Ca(OH)₂ in water: Ca(OH)₂ → Ca²⁺ + 2OH⁻. Thus, 1 mole of Ca(OH)₂ produces 2 moles of OH⁻ ions. moles of Ca(OH)₂ = 0.0105 M × 0.01 L = 1.05 × 10⁻⁴ moles of OH⁻ = 1.05 × 10⁻⁴ × 2 = 2.1 × 10⁻⁴
07

Calculate the concentration of OH⁻ ions

[OH⁻] = 2.1 × 10⁻⁴ mol / 0.5 L = 4.2 × 10⁻⁴ M
08

Calculate pH

pOH = -log(4.2 × 10⁻⁴) = 3.38 pH = 14 - 3.38 = 10.62 (c) [OH⁻] = 4.2 × 10⁻⁴ M; pH = 10.62 (d) A solution formed by mixing 20.0 mL of 0.015 M Ba(OH)₂ with 40.0 mL of 8.2 × 10⁻³ M NaOH.
09

Calculate the moles of OH⁻ ions

The dissociation of Ba(OH)₂: Ba(OH)₂ → Ba²⁺ + 2OH⁻. Thus, 1 mole of Ba(OH)₂ produces 2 moles of OH⁻ ions. moles of OH⁻ from Ba(OH)₂ = 0.015 M × 0.02 L × 2 = 6 × 10⁻⁴ mol moles of OH⁻ from NaOH = 8.2 × 10⁻³ M × 0.04 L = 3.28 × 10⁻⁴ Total moles of OH⁻ = 6 × 10⁻⁴ + 3.28 × 10⁻⁴ = 9.28 × 10⁻⁴ mol
10

Calculate the concentration of OH⁻ ions

Total volume = 20 mL + 40 mL = 60 mL = 0.06 L [OH⁻] = 9.28 × 10⁻⁴ mol / 0.06 L = 1.5467 × 10⁻² M
11

Calculate pH

pOH = -log(1.5467 × 10⁻²) = 1.81 pH = 14 - 1.81 = 12.19 (d) [OH⁻] = 1.5467 × 10⁻² M; pH = 12.19

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

Triethylamine, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{~N},\) has a \(\mathrm{p} K_{b}\) value of \(2.99 .\) Is triethylamine a stronger base than ammonia, \(\mathrm{NH}_{3} ?\)

(a) Give the conjugate base of the following Bronsted- Lowry acids: (i) \(\mathrm{HIO}_{3}\), (ii) \(\mathrm{NH}_{4}^{+}\). (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{O}^{2-}\) (ii) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)

If a substance is a Lewis acid, is it necessarily a Bronsted-Lowry acid? Is it necessarily an Arrhenius acid? Explain.

A \(0.100 M\) solution of bromoacetic acid \(\left(\mathrm{BrCH}_{2} \mathrm{COOH}\right)\) is \(13.2 \%\) ionized. Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{BrCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{BrCH}_{2} \mathrm{COOH}\right]\) and \(K_{a}\) for bromoacetic acid.

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{NO}_{2}^{-},\) (b) \(\mathrm{PO}_{4}^{3-}\) or \(\mathrm{AsO}_{4}^{3-},(\mathbf{c})\) \(\mathrm{HCO}_{3}^{-}\) or \(\mathrm{CO}_{3}^{2-}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

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.

Sign-up for free