Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 44

A 20.0 -mL sample of 0.150 \(\mathrm{M} \mathrm{KOH}\) is titrated with 0.125 \(\mathrm{M}\) \(\mathrm{HClO}_{4}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of acid have been added: (a) \(20.0 \mathrm{mL},\) (b) 23.0 \(\mathrm{mL}\) \((\mathbf{c}) 24.0 \mathrm{mL},(\mathbf{d}) 25.0 \mathrm{mL},(\mathbf{e}) 30.0 \mathrm{mL}\)

Short Answer

Expert verified
The pH values of the solution at each point during the titration with HClO₄ are: a) 11.70 b) 11.30 c) 7.00 d) 2.52 e) 2.12

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the reaction between KOH and HClO4 is: \(KOH(aq) + HClO_4(aq) \rightarrow KClO_4(aq) + H_2O(l)\)
02

Determine the moles of KOH and HClO4 added to the solution

For each volume of HClO4 (20.0 mL, 23.0 mL, 24.0 mL, 25.0 mL, and 30.0 mL), we will calculate the moles of KOH and HClO4: Initial moles of KOH in 20.0 mL solution = 0.150 M * 0.020 L = 0.0030 mol
03

Calculate moles of HClO4 added at each point

For each given volume of HClO4, the moles of HClO4 added can be calculated using the given concentration: a) 20.0 mL of HClO4: (0.125 M) * (0.020 L) = 0.0025 mol b) 23.0 mL of HClO4: (0.125 M) * (0.023 L) = 0.002875 mol c) 24.0 mL of HClO4: (0.125 M) * (0.024 L) = 0.0030 mol d) 25.0 mL of HClO4: (0.125 M) * (0.025 L) = 0.003125 mol e) 30.0 mL of HClO4: (0.125 M) * (0.030 L) = 0.00375 mol
04

Calculate moles of KOH and HClO4 at each point

We must take into account the moles reacting at each point, to decide if we must calculate the pH with the excess reactant, or the equilibrium of the solution: a) 20.0 mL of HClO4: Moles of \(KOH (0.0030 - 0.0025) = 0.0005\) mol, No leftover HClO4 b) 23.0 mL of HClO4: Moles of \(KOH (0.0030 - 0.002875) = 0.000125\) mol, No leftover HClO4 c) 24.0 mL of HClO4: At equivalence point, no leftover KOH or HClO4 d) 25.0 mL of HClO4: No leftover KOH, moles of \(HClO_4 (0.003125 - 0.0030) = 0.000125\) mol e) 30.0 mL of HClO4: No leftover KOH, moles of \(HClO_4 (0.00375 - 0.0030) = 0.00075\) mol
05

Calculate the pH at each point

To calculate the pH at each point, we need to find the concentration of H3O+ or OH- and then use the pH or pOH formula. If there's excess KOH, calculate the pOH and then find the pH. If there's excess HClO4, calculate the pH directly. a) 20.0 mL of HClO4: Excess KOH, pOH = -log10([OH-]) = -log10(0.0005 mol / 0.040 L) = 2.30; pH = 14 - pOH = 11.70 b) 23.0 mL of HClO4: Excess KOH, pOH = -log10([OH-]) = -log10(0.000125 mol / 0.043 L) = 2.70; pH = 14 - pOH = 11.30 c) 24.0 mL of HClO4: Equivalence point, pH = 7 d) 25.0 mL of HClO4: Excess HClO4, pH = -log10([H3O+]) = -log10(0.000125 mol / 0.045 L) = 2.52 e) 30.0 mL of HClO4: Excess HClO4, pH = -log10([H3O+]) = -log10(0.00075 mol / 0.050 L) = 2.12 In conclusion, the pH values are as follows: a) 11.70 b) 11.30 c) 7.00 d) 2.52 e) 2.12

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

Lead(II) carbonate, PbCO \(_{3},\) is one of the components of the passivating layer that forms inside lead pipes.(a) If the \(K_{s p}\) for \(\mathrm{PbCO}_{3}\) is \(7.4 \times 10^{-14}\) what is the molarity of \(\mathrm{Pb}^{2+}\) in a saturated solution of lead(II) carbonate? (b) What is the concentration in ppb of \(\mathrm{Pb}^{2+}\) ions in a saturated solution? (c) Will the solubility of \(\mathrm{PbCO}_{3}\) increase or decrease as the \(\mathrm{pH}\) is lowered? \((\boldsymbol{d} )\)The EPA threshold for acceptable levels of lead ions in water is 15 ppb. Does a saturated solution of lead(II) carbonate produce a solution that exceeds the EPA limit?

The solubility of two slightly soluble salts of \(\mathrm{M}^{2+}, \mathrm{MA}\) and \(\mathrm{MZ}_{2},\) is the same, \(4 \times 10^{-4} \mathrm{mol} / \mathrm{L}\) . (a) Which has the larger numerical value for the solubility product constant? (b) In a saturated solution of each salt in water, which has the higher concentration of \(\mathrm{M}^{2+} ?\) (c) If you added an equal volume of a solution saturated in MA to one saturated in \(\mathrm{MZ}_{2},\) what would be the equilibrium concentration of the cation, \(\mathrm{M}^{2+}\) ?

How many microliters of 1.000\(M\) NaOH solution must be added to 25.00 \(\mathrm{mL}\) of a 0.1000 \(\mathrm{M}\) solution of lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\) or \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to produce a buffer with \(\mathrm{pH}=3.75 ?\)

A solution contains \(2.0 \times 10^{-4} M \mathrm{Ag}^{+}(a q)\) a n d \(1.5 \times 10^{-3} M \mathrm{Pb}^{2+}(a q)\). I f N a I i s a d d e d , w i l l A g I (\(K_{s p}=8.3 \times 10^{-17}\)) or \(\mathrm{PbI}_{2}\left(K_{s p}=7.9 \times 10^{-9}\right)\) precipitate first? Specify the concentration of \(\mathrm{I}^{-}(a q)\) needed to begin precipitation.

The solubility of \(\mathrm{CaCO}_{3}\) is pH dependent. (a) Calculate the molar solubility of \(\mathrm{CaCO}_{3}\left(K_{s p}=4.5 \times 10^{-9}\right)\) neglecting the acid-base character of the carbonate ion. (b) Use the \(K_{b}\) expression for the \(\mathrm{CO}_{3}^{2-}\) ion to determine the equilibrium constant for the reaction $$\mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons_{\mathrm{Ca}^{2+}(a q)+\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)}$$ (c) If we assume that the only sources of \(\mathrm{Ca}^{2+}, \mathrm{HCO}_{3}^{-}\) and \(\mathrm{OH}^{-}\) ions are from the dissolution of \(\mathrm{CaCO}_{3},\) what is the molar solubility of \(\mathrm{CaCO}_{3}\) using the equilibrium expression from part (b)? \((\boldsymbol{d} )\)What is the molar solubility of \(\mathrm{CaCO}_{3}\) at the pH of the ocean \((8.3) ?(\mathbf{e})\) If the \(\mathrm{pH}\) is buffered at \(7.5,\) what is the molar solubility of \(\mathrm{CaCO}_{3} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

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