Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 92

Mathematically prove that the \(\mathrm{pH}\) at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to $\mathrm{p} K_{a}$ for the acid.

Short Answer

Expert verified
At the halfway point of titration of a weak acid with a strong base, the remaining concentration of the weak acid HA is \(\frac{1}{2}C_{A}\), and the concentration of the conjugate base MA is also \(\frac{1}{2}C_{A}\), since the stoichiometry of the reaction is 1:1. The equilibrium constant, \(K_{a}\), can be expressed as \(K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}\). At the halfway point of titration, the pH is calculated as \(\mathrm{pH} = -\log(K_{a})\), and since p\(K_{a} = -\log(K_{a})\), the pH at the halfway point is equal to the p\(K_{a}\) for the weak acid.

Step by step solution

01

Write the chemical reaction

For the titration of a weak acid HA with a strong base MOH, the chemical reaction can be written as: \[ HA + MOH \rightarrow MA + H_{2}O \] At the halfway point of the titration, exactly half of the weak acid has been neutralized by the strong base.
02

Determine the relevant concentrations at the halfway point

Let the initial concentration of the weak acid HA be denoted by \(C_{A}\), and let the volume of the MOH solution used to reach the halfway point be \(V_{1}\). Since half of the acid has been neutralized, the remaining concentration of HA is \(\frac{1}{2}C_{A}\), and the concentration of the conjugate base MA is also \(\frac{1}{2}C_{A}\), since the stoichiometry of the reaction is 1:1.
03

Write an expression for the equilibrium concentrations

Since the weak acid is incompletely ionized in water, we can write the ionization equilibrium as: \[ HA \rightleftharpoons H^{+} + A^{-} \] Then, the equilibrium constant, \(K_{a}\), can be expressed as: \[ K_{a} = \frac{[H^{+}][A^{-}]}{[HA]} \] At the halfway point, \([A^{-}] = \frac{1}{2}C_{A}\) and \([HA] = \frac{1}{2}C_{A}\).
04

Solve for [H+] (hydrogen ion concentration)

Substitute the concentrations of \(A^{-}\) and \(HA\) at the halfway point into the equilibrium constant expression: \[ K_{a} = \frac{[H^{+}]\left(\frac{1}{2}C_{A}\right)}{\left(\frac{1}{2}C_{A}\right)} \] Since we have equal concentrations of weak acid HA and its conjugate base A-, we can simplify this expression as follows: \[ K_{a} = [H^{+}] \]
05

Calculate the pH at the halfway point

Now that we have an expression for the hydrogen ion concentration, we can calculate the \(\mathrm{pH}\) at the halfway point of the titration using the definition of \(\mathrm{pH}\): \[ \mathrm{pH} = -\log ([H^{+}]) \] At the halfway point, \([H^{+}] = K_{a}\), so we can substitute this into the \(\mathrm{pH}\) equation: \[ \mathrm{pH} = -\log(K_{a}) \] By definition, p\(K_{a} = -\log(K_{a})\), so we can rewrite the equation as: \[ \mathrm{pH} = \mathrm{p} K_{a} \] Thus, at the halfway point of titration of a weak acid with a strong base, the \(\mathrm{pH}\) is equal to the p\(K_{a}\) for the weak acid.

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

Suppose you want to do a physiological experiment that calls for a pH 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid $\mathrm{H}_{2} \mathrm{~A}\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)$ or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and a 1.0 \(M\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH}\) 6.50? (Ignore any volume change.)

Fluoridation of drinking water is employed in many places to aid in the prevention of tooth decay. Typically. the \(\mathrm{F}^{-}\) ion concentration is adjusted to about \(1 \mathrm{ppm}\). Some water supplies are also "hard"; that is, they contain certain cations such as \(\mathrm{Ca}^{2+}\) that interfere with the action of soap. Consider a case where the concentration of \(\mathrm{Ca}^{2+}\) is \(8 \mathrm{ppm}\). Could a precipitate of \(\mathrm{CaF}_{2}\) form under these conditions? (Make any necessary approximations.)

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 pH of the solution increase, decrease, or stay the same?

Consider a beaker containing a saturated solution of \(\mathrm{PbI}_{2}\) in equilibrium with undissolved \(\mathrm{PbI}_{2}(s) .\) Now solid KI 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?

A solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added dropwise to a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Ba}^{2+}(a q)\) and $0.010 \mathrm{M}\( in \)\mathrm{Sr}^{2+}(a q) .(\mathbf{a}) \mathrm{What}$ concentration of \(\mathrm{SO}_{4}^{2-}\) is necessary to begin precipitation? (Neglect volume changes. $\mathrm{BaSO}_{4}: K_{s p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}:\( \)K_{s p}=3.2 \times 10^{-7} .$ ) (b) Which cation precipitates first? (c) What is the concentration of $\mathrm{SO}_{4}^{2-}(a q)$ when the second cation begins to precipitate?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

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