Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

How many grams of \(\mathrm{NaOH}\) should be added in \(500 \mathrm{ml}\) of \(2 \mathrm{M}\) acetic acid solution to get a buffer solution of maximum buffer capacity? (a) \(20.0\) (b) \(10.0\) (c) \(40.0\) (d) \(30.0\)

Short Answer

Expert verified
20.0 grams of NaOH should be added to 500 ml of 2 M acetic acid solution to get a buffer solution of maximum buffer capacity.

Step by step solution

01

Understanding Maximum Buffer Capacity

Maximum buffer capacity occurs when the pH of the solution is equal to the pKa of the acid in the buffer. For acetic acid, the pKa is approximately 4.75. At maximum buffer capacity, the concentrations of the acid (acetic acid) and its conjugate base (acetate ion) are equal.
02

Calculate Moles of Acetic Acid

Firstly, calculate the moles of acetic acid in the solution using the concentration and the volume. Since the solution concentration is 2 M (moles per liter), and the volume is 500 ml (0.5 liters), the moles of acetic acid are calculated as follows: Moles = Molarity × Volume = 2 M × 0.5 L = 1 mole.
03

Determining Moles of NaOH Needed

Since the acetic acid needs to be present in equal amounts to its conjugate base for maximum buffer capacity, we will need enough NaOH to convert half of the acetic acid to its conjugate base. Moles of NaOH needed = 1 mole / 2 = 0.5 moles.
04

Convert Moles of NaOH to Grams

The molar mass of NaOH is approximately 40 g/mol. To find the mass of NaOH needed, multiply the moles needed by the molar mass: Mass = Moles × Molar Mass = 0.5 moles × 40 g/mol = 20 grams.

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 Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Maximum Buffer Capacity
One of the essential concepts in buffer solution chemistry is the 'maximum buffer capacity'. This occurs when a buffer can effectively neutralize added acids or bases, achieving the highest resistance to changes in pH levels. To attain this condition, the pH of the buffer solution should be identical to the pKa of the buffering acid. For acetic acid, the pKa is around 4.75. This translates into the need for equimolar amounts of the weak acid and its conjugate base in the solution.

In the context of the given exercise, where acetic acid is used, achieving maximum buffer capacity involves adding a precise quantity of a strong base, NaOH in this case, to convert exactly half of the acetic acid molecules to acetate ions. This equal concentration of acetic acid and acetate ions creates a buffer system robust enough to resist pH changes, upholding the properties of the acetic acid's pKa value.
Acid-Base Equilibria
Acid-base equilibria is a fundamental chemical principle that describes how acids and bases interact in solution. It plays a critical role in understanding buffer solutions. Buffer solutions consist of a weak acid and its conjugate base (or a weak base and its conjugate acid) and maintain a stable pH upon the addition of small quantities of acids or bases.

The acetic acid and sodium hydroxide (NaOH) interaction in the exercise exemplifies an acid-base reaction that leads to an equilibrium. During the reaction, NaOH (a strong base) reacts with acetic acid (a weak acid) to produce its conjugate base, acetate. The solution's ability to resist pH changes hinges on the presence of both the acetic acid and its conjugate base at equilibrium. This characteristic supports the effectiveness of a buffer when the system keeps the concentrations of the acid and its conjugate base at equilibrium — a scenario that occurs when the solution reaches its maximum buffer capacity.
Stoichiometry Calculations
Stoichiometry calculations are the mathematical methods used to relate quantities of reactants and products in chemical reactions. These calculations are vital for preparing buffer solutions with the desired properties. In this case, the stoichiometry of the reaction between acetic acid and NaOH tells us that they react in a 1:1 ratio to form water and acetate ions.

With this information, we can determine the amount of NaOH needed to react with half of the acetic acid present, which ensures a buffer of maximum capacity. By using the molarity of the acetic acid solution and the volume provided, we calculate the moles of acetic acid. We then use the 1:1 reaction ratio to find out that we need half as many moles of NaOH. Finally, we convert these moles to grams by using the molar mass of NaOH. The step-by-step solution provided reveals that diligent, systematic stoichiometry is crucial to deduce the right amount of NaOH needed to create a buffer solution at its maximum buffering capacity.

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

What is the aqueous ammonia concentration of a solution prepared by dissolving \(0.15\) mole of \(\mathrm{NH}_{4}^{+} \mathrm{CH}_{3} \mathrm{COO}^{-}\) in 1 L of water? Given: \(K_{\text {a }}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) \(=1.8 \times 10^{-5} ; K_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=1.8 \times 10^{-5}\) (a) \(8.3 \times 10^{-4} \mathrm{M}\) (b) \(0.15 \mathrm{M}\) (c) \(5.52 \times 10^{-3} \mathrm{M}\) (d) \(3.8 \times 10^{-4} \mathrm{M}\)

Following five solutions of \(\mathrm{KOH}\) were prepared as: first, \(0.1\) mole in \(1 \mathrm{~L} ;\) second. \(0.2\) mole in \(2 \mathrm{~L}\); third, \(0.3\) mole in \(3 \mathrm{~L}\); fourth, \(0.4\) mole in \(4 \mathrm{~L} ;\) fifth, \(0.5\) mole in \(5 \mathrm{~L}\). The \(\mathrm{pH}\) of resultant solution, when all these solutions are mixed, is (a) 2 (b) 1 (c) 13 (d) 7

The solubility product of \(\mathrm{Co}(\mathrm{OH})_{3}\) is \(2.7 \times 10^{-43}\). The pH of saturated solution of \(\mathrm{Co}(\mathrm{OH})_{3}\) is about (a) \(7.0\) (b) \(11.0\) (c) \(3.0\) (d) \(3.48\)

The solubility product of \(\mathrm{CaF}_{2}\) is \(1.08\) \(\times 10^{-10}\). What mass of \(\mathrm{CaF}_{2}\) will dissolve in \(500 \mathrm{ml}\) water in order to make a saturated solution? \((\mathrm{Ca}=40, \mathrm{~F}=19)\) (a) \(3 \times 10^{-4} \mathrm{~g}\) (b) \(1.17 \times 10^{-2} \mathrm{~g}\) (c) \(1.17 \mathrm{mg}\) (d) \(3 \times 10^{-3} \mathrm{~g}\)

The dissociation constant of formic acid is \(0.00024\). The hydrogen ion concentration in \(0.002 \mathrm{M}-\mathrm{HCOOH}\) solution is nearly (a) \(6.93 \times 10^{-4} \mathrm{M}\) (b) \(4.8 \times 10^{-7} \mathrm{M}\) (c) \(5.8 \times 10^{-4} \mathrm{M}\) (d) \(1.4 \times 10^{-4} \mathrm{M}\)

See all solutions

Recommended explanations on Chemistry Textbooks

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