Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 48

Calculate the \(\mathrm{pH}\) at the equivalence point in titrating 0.100 M solutions of each of the following with 0.080 \(\mathrm{M}\) NaOH: (a) hydrobromic acid (HBr), (b) chlorous acid (HClO_{2} ) , (c) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\)

Short Answer

Expert verified
The pH at the equivalence points for each titration are: (a) Hydrobromic acid (HBr): pH = 7, (b) Chlorous acid (HClO₂): pH ≈ 6.96, and (c) Benzoic acid (C₆H₅COOH): pH ≈ 8.63.

Step by step solution

01

Write the balanced equation for the reaction.

The reaction between HBr (a strong acid) and NaOH (a strong base) can be written as: HBr + NaOH -> NaBr + H₂O
02

Calculate the moles of reactants at the equivalence point

At the equivalence point, both reactants are completely consumed: moles_HBr = moles_NaOH Let V (in liters) be the volume of NaOH required to reach the equivalence point: 0.100 M * V = 0.080 M * V The volume, V, is not needed as it cancels out as they are equal.
03

Calculate the pH

Since both the acid and the base are strong, the resulting solution will be neutral (pH = 7) at the equivalence point. #b) Chlorous acid (HClO₂)#:
04

Write the balanced equation for the reaction.

The reaction between HClO₂ (a weak acid) and NaOH (a strong base) can be written as: HClO₂ + NaOH -> NaClO₂ + H₂O
05

Calculate the moles of reactants at the equivalence point.

At the equivalence point, both reactants are completely consumed: moles_HClO₂ = moles_NaOH Let V₂ (in liters) be the volume of NaOH required to reach the equivalence point: 0.100 M * V₂ = 0.080 M * V₂ The volume, V₂, is not needed as it cancels out as they are equal.
06

Calculate the concentration of ClO₂- (A-) at the equivalence point.

Since at equivalence point, [HClO₂] = [NaOH], so the concentration of ClO₂- will also be equal to the initial concentration of HClO₂: [A-] = 0.100 M
07

Use Kb for ClO₂- to find the hydroxide ion concentration ([OH⁻]).

The Kb for ClO₂- can be found using the Ka relationship: Kb = Kw / Ka Where Kw = 1.0 × 10-14 and Ka = 1.2 x 10^{-2}. Kb = (1.0 × 10-14) / (1.2 × 10^{-2}) Solving for Kb, we get: Kb = 8.33 × 10^{-13} Now we make an ICE table for ClO₂-, knowing that the initial concentration of OH⁻ is 0: Initial: [ClO₂-] = 0.100 M | [OH⁻] = 0 Change: [ClO₂-] = -x | [OH⁻] = x Equilibrium: [ClO₂-] = 0.100-x | [OH⁻] = x Now, as Kb is very small, we can assume that x will be very small and 0.100 - x will be approximately equal to 0.100: Kb = (x)(x) / 0.100 Solving for x, we get: x = [OH⁻] = 9.13 × 10^{-8} M
08

Calculate the pH.

Since we have calculated the [OH⁻], we can now find the pOH using the following equation: pOH = -log([OH⁻]) pOH = -log(9.13 × 10^{-8}) pOH ≈ 7.04 Finally, using the relationship between pH and pOH: pH = 14 - pOH pH ≈ 6.96 #c) Benzoic acid (C₆H₅COOH)#: Repeat steps 1-5 from case (b), but instead of HClO₂, use C₆H₅COOH with Ka = 6.3 × 10^{-5}. After performing the calculations, we get: pH ≈ 8.63 Thus, the pH at the equivalence points for each titration are: (a) HBr: pH = 7, (b) HClO₂: pH ≈ 6.96, and (c) C₆H₅COOH: pH ≈ 8.63.

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

Consider a beaker containing a saturated solution of \(\mathrm{PbI}_{2}\) in equilibrium with undissolved \(\mathrm{PbI}_{2}(s)\). Now solid \(\mathrm{K} \mathrm{I}\) is added to this solution. (a) Will the amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Pb}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of \(I\) ions in solution increase or decrease?

How many milliliters of 0.0850\(M \mathrm{NaOH}\) are required to titrate each of the following solutions to the equivalence point: (a) 40.0 \(\mathrm{mL}\) of \(0.0900 \mathrm{M} \mathrm{HNO}_{3},\) (\mathbf{b} ) 35.0 \(\mathrm{mL}\) of \(0.0850 M \mathrm{CH}_{3} \mathrm{COOH},(\mathbf{c}) 50.0 \mathrm{mL}\) of a solution that contains 1.85 \(\mathrm{g}\) of \(\mathrm{HCl}\) per liter?

(a) Will \(\mathrm{Co}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathbf{p H}\) of a 0.020 M solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) is adjusted to 8.5? (b) Will \(\mathrm{AgIO}_{3}\) precipitate when 20 mL of 0.010 M \(\mathrm{AglO}_{3}\) is mixed with 10 mL of 0.015 \(M \mathrm{NaIO}_{3}\)? ( \(K_{s p}\) of \(\mathrm{AgIO}_{3}\) is \(3.1 \times 10^{8}\))?

Assume that 30.0 \(\mathrm{mL}\) of a 0.10 \(\mathrm{M}\) solution of a weak base \(\mathrm{B}\) that accepts one proton is titrated with a 0.10\(M\) solution of the monoprotic strong acid HA. (a) How many moles of HA have been added at the equivalence point? (b) What is the predominant form of B at the equivalence point? (a) Is the pH \(7,\) less than \(7,\) or more than 7 at the equivalence point?\( (\mathbf{d} )\) Which indicator, phenolphthalein or methyl red, is likely to be the better choice for this titration?

Consider the equilibrium $$\mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ Suppose that a salt of \(\mathrm{HB}^{+}(a q)\) is added to a solution of \(\mathrm{B}(a q)\) at equilibrium. (a) Will the equilibrium constant for the reaction increase, decrease, or stay the same? (b) Will the concentration of \(\mathrm{B}(a q)\) increase, decrease, or stay the same? (c) Will the \(\mathrm{pH}\) of the solution increase, decrease, or stay the same?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

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.

Sign-up for free