Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 174

Making use of the assumptions we ordinarily make in calculating the \(\mathrm{pH}\) of an aqueous solution of a weak acid, calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-6}-M\) solution of hypobromous acid (HBrO, \(K_{\mathrm{a}}=2 \times 10^{-9}\) ). What is wrong with your answer? Why is it wrong? Without trying to solve the problem, explain what has to be included to solve the problem correctly.

Short Answer

Expert verified
The calculated pH of a $1.0 \times 10^{-6}-M$ solution of hypobromous acid (HBrO) with $K_{a}=2 \times 10^{-9}$ is approximately 7.85. However, this is incorrect because it neglects the autoionization of water and its contribution to the [H+] concentration. To solve the problem correctly, we should take into account the autoionization of water (Kw = 1 x 10^{-14}) and add that to the [H+] due to HBrO ionization before calculating the pH.

Step by step solution

01

Write the weak acid ionization equation.

The ionization equation for hypobromous acid (HBrO) in water is: HBrO (aq) <=> H+ (aq) + BrO- (aq).
02

Write the expression for Ka.

The expression for Ka (acid dissociation constant) for HBrO is: Ka = \(\frac{[H^+][BrO^-]}{[HBrO]}\)
03

Set up initial and equilibrium concentrations molecule table (ICE table).

An ICE (Initial, Change, Equilibrium) table is used to set up the initial, changes that occur, and the equilibrium concentrations for all components of the ionization process. Initial concentrations of the components are: [HBrO] = 1.0 × 10^{-6} M [H+] = 0 M (assuming no dissociation initially) [BrO-] = 0 M (assuming no dissociation initially) During the ionization process, let 'x' be the change in concentration of H+. We then have the change in the concentrations as: [HBrO] -x [H+] +x [BrO-] +x Equilibrium concentrations are: [HBrO] = (1.0 × 10^{-6} - x) M [H+] = x M [BrO-] = x M
04

Substitute equilibrium concentrations into Ka expression and solve for x.

Plug the equilibrium concentrations of HBrO, H+, and BrO- into the Ka expression: Ka = 2 × 10^{-9} = \(\frac{x^2}{(1.0 × 10^{-6} - x)}\) Now, we assume that x is much smaller than the initial concentration of HBrO, so we simplify the equation as: 2 × 10^{-9} = \(\frac{x^2}{1.0 × 10^{-6}}\) Solve for x: x = \(\sqrt{(2 × 10^{-9})(1.0 × 10^{-6})}\) x = \(\sqrt{2 × 10^{-15}}\) x ≈ 1.4 × 10^{-8} M
05

Calculate the pH.

Since [H+] ≈ x, and pH = -log([H+]), therefore: pH = -log(1.4 × 10^{-8}) pH ≈ 7.85
06

What is wrong with our answer?

The assumption that the concentration of [H+] ions is solely due to the ionization of HBrO is wrong, as it neglects the autoionization of water. The [H+] ion concentration from water has to be taken into account since the HBrO concentration is very close to water autoionization equilibrium concentration.
07

How to solve the problem correctly?

To correctly calculate the pH value in this situation, we should take into account the [H+] contribution from the autoionization of water (Kw = 1 x 10^{-14}) and add that to the [H+] due to HBrO ionization before calculating the pH. This ensures that we consider both sources of H+ ions and do not underestimate the actual pH value of the solution.

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

Isocyanic acid (HNCO) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation \(2 \mathrm{NaOCN}(s)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \longrightarrow 2 \mathrm{HNCO}(l)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s)\).

Calculate \(\left[\mathrm{CO}_{3}^{2-}\right]\) in a \(0.010-M\) solution of \(\mathrm{CO}_{2}\) in water (usually written as \(\mathrm{H}_{2} \mathrm{CO}_{3}\) ). If all the \(\mathrm{CO}_{3}{ }^{2-}\) in this solution comes from the reaction $$\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q)$$ what percentage of the \(\mathrm{H}^{+}\) ions in the solution is a result of the dissociation of \(\mathrm{HCO}_{3}^{-} ?\) When acid is added to a solution of sodium hydrogen carbonate \(\left(\mathrm{NaHCO}_{3}\right)\), vigorous bubbling occurs. How is this reaction related to the existence of carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\) molecules in aqueous solution?

Use Table \(14.3\) to help order the following bases from strongest to weakest. $$ \mathrm{NO}_{3}^{-}, \quad \mathrm{H}_{2} \mathrm{O}, \quad \mathrm{NH}_{3}, \quad \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N} $$

Calculate the mass of \(\mathrm{HONH}_{2}\) required to dissolve in enough water to make \(250.0 \mathrm{~mL}\) of solution having a \(\mathrm{pH}\) of \(10.00\left(K_{\mathrm{b}}=\right.\) \(\left.1.1 \times 10^{-8}\right)\).

Calculate \(\left[\mathrm{OH}^{-}\right]\) in a \(3.0 \times 10^{-7}-M\) solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

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