/*! 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 76 Calculate the molar concentratio... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 76

Calculate the molar concentration of \(\mathrm{OH}^{-}\) ions in a \(0.550 \mathrm{M}\) solution of hypobromite ion \(\left(\mathrm{BrO}^{-} ;\right.\) \(K_{b}=4.0 \times 10^{-6}\) ). What is the \(\mathrm{pH}\) of this solution?

Short Answer

Expert verified
The equilibrium concentration of \(\mathrm{OH}^{-}\) ions in the \(0.550\mathrm{M}\) solution of hypobromite ion is approximately \(2.98 \times 10^{-3} \mathrm{M}\), and the corresponding \(\mathrm{pH}\) of the solution is approximately \(11.47\).

Step by step solution

01

Write the expression for the base dissociation reaction

For the hypobromite ion, the base dissociation reaction is: \[\mathrm{BrO}^{-} + \mathrm{H_2O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{HBrO}\]
02

Write the expression for the base dissociation constant \(K_b\)

The expression for the base dissociation constant \(K_b\) is: \[K_b = \frac{[\mathrm{OH}^{-}][\mathrm{HBrO}]}{[\mathrm{BrO}^{-}]}\]
03

Set up an ICE table for the reaction

An ICE table will help us determine the equilibrium concentrations of the species involved in the reaction. The initial concentration of \(\mathrm{BrO}^{-}\) is \(0.550 \mathrm{M}\), while the initial concentrations of \(\mathrm{OH}^{-}\) and \(\mathrm{HBrO}\) are both \(0\). ``` [BrO鈦籡 [OH鈦籡 [HBrO] I 0.550 M 0 0 C -x +x +x E 0.550-x M x x ```
04

Solve for the equilibrium concentration of \(\mathrm{OH}^{-}\) ions

At equilibrium, we can plug the ICE table values into the \(K_b\) expression: \[4.0 \times 10^{-6} = \frac{x \cdot x}{0.550 - x}\] Assuming that x is small compared to \(0.550\), we can approximate the denominator to be \(0.550\): \[4.0 \times 10^{-6} = \frac{x^2}{0.550}\] Now, solve for x (the equilibrium concentration of \(\mathrm{OH}^{-}\) ions): \[x^2 = 4.0 \times 10^{-6} \cdot 0.550\] \[x = \sqrt{4.0 \times 10^{-6} \cdot 0.550} \approx 2.98 \times 10^{-3} \mathrm{M}\]
05

Calculate \(\mathrm{pOH}\) and \(\mathrm{pH}\) of the solution

Now that we have the equilibrium concentration of \(\mathrm{OH}^{-}\) ions, we can calculate the \(\mathrm{pOH}\) and \(\mathrm{pH}\) of the solution: \[\mathrm{pOH} = -\log{[\mathrm{OH}^{-}]} = -\log{(2.98 \times 10^{-3})} \approx 2.53\] Recall that \(\mathrm{pH} + \mathrm{pOH} = 14\), so we can find the \(\mathrm{pH}\) of the solution: \[\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 2.53 \approx 11.47\] Thus, the \(\mathrm{pH}\) of the \(0.550 \mathrm{M}\) solution of hypobromite ion is approximately \(11.47\).

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!

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

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\) (b) \(2.250 \mathrm{~g}\) of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution, \((c) 100 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}\) diluted to \(2.00 \mathrm{~L},(\mathrm{~d})\) a solution formed by adding \(5.00 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{KOH}\) to \(15.0 \mathrm{~mL}\) of \(9.5 \times 10^{-2} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\)

Saccharin, a sugar substitute, is a weak acid with \(\mathrm{P} K_{a}=2.32\) at \(25^{\circ} \mathrm{C}\). It ionizes in aqueous solution as follows: $$ \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q)=\mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q) $$ What is the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of this substance?

(a) Which of the following is the stronger BronstedLowry acid, \(\mathrm{HBrO}\) or \(\mathrm{HBr} ?(\mathrm{~b})\) Which is the stronger Brensted-Lowry base, \(\mathrm{F}^{-}\) or \(\mathrm{Cl}^{-}\) ? Briefly explain your choices.

Which member of each pair produces the more acidic aqueous solution: (a) \(\mathrm{ZnBr}_{2}\) or \(\mathrm{CdCl}_{2}\), (b) \(\mathrm{CuCl}\) or \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\), (c) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) or \(\mathrm{NiBr}_{2}\) ? Explain.

The iodate ion is reduced by sulfite according to the following reaction: $$ \mathrm{IO}_{3}^{-}(a q)+3 \mathrm{SO}_{3}{ }^{2-}(a q) \longrightarrow \mathrm{I}^{-}(a q)+3 \mathrm{SO}_{4}{ }^{2-}(a q) $$ The rate of this reaction is found to be first order in \(\mathrm{IO}_{3}\). first order in \(\mathrm{SO}_{3}{ }^{2-}\), and first order in \(\mathrm{H}^{+}\). (a) Write the rate law for the reaction. (b) By what factor will the rate of the reaction change if the \(\mathrm{pH}\) is lowered from \(5.00\) to 3.50? Does the reaction proceed faster or slower at the lower pH? (c) By using the concepts discussed in Section 14.6, explain how the reaction can be pH- dependent
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Nuclear 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.