Logout Open In App
Open In App
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

Chapter 7: Problem 15

What mass of NaOH should be dissolved in sufficient water to get \(20 \mathrm{~m}^{3}\) of an aqueous solution of \(\mathrm{pH}, 7.3\), at \(25^{\circ} \mathrm{C}\) ? (a) \(0.16 \mathrm{~g}\) (b) \(1.6 \times 10^{-4} \mathrm{~g}\) (c) \(0.04 \mathrm{~g}\) (d) \(0.12 \mathrm{~g}\)

Short Answer

Expert verified
The mass of NaOH needed is approximately 0.16 g.

Step by step solution

01

- Understanding pH and pOH

pH is a measure of the hydrogen ion concentration in a solution, with the formula pH = -log[H+]. A pH of 7.3 indicates an alkaline solution. Since the pH and pOH of a solution add up to 14 at room temperature, calculate pOH using pOH = 14 - pH.
02

- Calculating [OH-] concentration

Using the formula pOH = -log[OH-], rearrange the equation to solve for the hydroxide ion concentration, [OH-] = 10^(-pOH).
03

- Converting to moles of NaOH

Knowing that each NaOH molecule releases one hydroxide ion (OH-) in solution, the concentration of NaOH will be equal to the [OH-] calculated. To calculate moles of NaOH, use the volume of the solution and multiply by the [OH-].
04

- Finding the mass of NaOH

The molar mass of NaOH is 40 g/mol. Use this to convert the moles of NaOH to grams.
05

- Applying significant figures

Apply the appropriate number of significant figures to your final answer based on the initial pH value to find the answer.

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!

Key Concepts

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

Hydroxide Ion Concentration
Understanding hydroxide ion concentration is fundamental to solving a wide range of chemistry problems, particularly when working with pH calculations and basic solutions. To clarify, the hydroxide ion concentration, denoted as [OH-], represents the amount of hydroxide ions present in a solution.

When a substance like NaOH (sodium hydroxide) dissolves in water, it dissociates into Na+ (sodium ions) and OH- (hydroxide ions). Since NaOH is a strong base, it completely dissociates, meaning the concentration of OH- ions in the solution is equal to the concentration of NaOH dissolved.

To find the hydroxide ion concentration from a given pOH, you can use the formula \[ [OH^-] = 10^{(-pOH)} \]. By inserting the value of pOH into this equation, you can determine the [OH-] in moles per liter (M), which is crucial for further calculations, such as determining the amount of a base required to achieve a certain pH, as in our example problem.
pOH and pH Relationship
The pOH and pH values are interconnected in a way that is essential for understanding the acidity or basicity of a solution. At a temperature of 25 degrees Celsius, the sum of the pH and pOH of an aqueous solution always equals 14.

Using the relationship \[ pOH = 14 - pH \], we can calculate the pOH given a pH, and vice versa. For instance, in our exercise with a solution of pH 7.3, the pOH can be calculated as \[ pOH = 14 - 7.3 = 6.7 \].

This pOH value can then be used to determine the hydroxide ion concentration, which is particularly useful when the solution in question is basic, as in the case of a NaOH solution. Knowing the pOH and consequently the [OH-] allows us to calculate the amount of base needed to achieve the desired pH.
Moles and Molar Mass of NaOH
To carry out pH calculations involving NaOH, it is essential to have a grasp on the concepts of moles and molar mass. The molar mass of a substance is the mass of one mole of its particles, often expressed in grams per mole (g/mol).

For sodium hydroxide (NaOH), the molar mass is 40 g/mol. This value is crucial when converting between the mass of NaOH and the number of moles needed for a particular concentration. In the context of the exercise, knowing that the [OH-] corresponds to the concentration of NaOH in the solution, we can calculate the moles of NaOH by multiplying the [OH-] by the volume of the solution in liters.

Finally, by multiplying the number of moles of NaOH by its molar mass, we can determine the mass in grams that must be dissolved in water to achieve the target pH: \[ \text{Mass of NaOH (g)} = \text{Moles of NaOH} \times \text{Molar Mass of NaOH (g/mol)} \]. Using this process, errors can be minimized and the exact mass required can be more accurately determined.

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

The addition of sodium acetate to acetic acid solution will cause (a) increase in its \(\mathrm{pH}\) value (b) decrease in its \(\mathrm{pH}\) value (c) no change in \(\mathrm{pH}\) value (d) change in \(\mathrm{pH}\) which cannot be predicted

Solid \(\mathrm{BaF}_{2}\) is added to a solution containing \(0.1\) mole of \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) solution (1 L) until equilibrium is reached. If the \(K_{\text {sp }}\) of \(\mathrm{BaF}_{2}\) and \(\mathrm{BaC}_{2} \mathrm{O}_{4}\) is \(10^{-6} \mathrm{~mol}^{3} \mathrm{~L}^{-3}\) and \(10^{-10} \mathrm{~mol}^{2} \mathrm{~L}^{-2}\), respectively, find the equilibrium concentration of \(\mathrm{Ba}^{2+}\) in the solution. Assume addition of \(\mathrm{BaF}_{2}\) does not cause any change in volume. (a) \(0.2 \mathrm{M}\) (b) \(4 \times 10^{-6} \mathrm{M}\) (c) \(2.5 \times 10^{-5} \mathrm{M}\) (d) \(2.5 \times 10^{-6} \mathrm{M}\)

A volume of \(10 \mathrm{ml}\) of \(0.1 \mathrm{M}\) tribasic acid, \(\mathrm{H}_{3} \mathrm{~A}\) is titrated with \(0.1 \mathrm{M}-\mathrm{NaOH}\) solution. What is the ratio (approximate value) of \(\frac{\left[\mathrm{H}_{3} \mathrm{~A}\right]}{\left[\mathrm{A}^{3-}\right]}\) at the second equivalent point? Given: \(K_{1}=7.5 \times 10^{-4} ; K_{2}=10^{-8}\); \(K_{3}=10^{-12}\) (a) \(10^{-4}\) (b) \(10^{-3}\) (c) \(10^{-7}\) (d) \(10^{-6}\)

When \(20 \mathrm{ml}\) of \(0.2 \mathrm{M}-\mathrm{DCl}\) solution is mixed with \(80 \mathrm{ml}\) of \(0.1 \mathrm{M}-\mathrm{NaOD}\) solution, \(\mathrm{pD}\) of the resulting solution becomes \(13.6 .\) The ionic product of heavy water, \(\mathrm{D}_{2} \mathrm{O}\), is (a) \(10^{-15}\) (b) \(10^{-16}\) (c) \(4 \times 10^{-15}\) (d) \(4 \times 10^{-16}\)

Calculate \(\left[\mathrm{S}^{2}\right]\) in a solution originally having \(0.1 \mathrm{M}-\mathrm{HCl}\) and \(0.2 \mathrm{M}-\mathrm{H}_{2} \mathrm{~S}\). For \(\mathrm{H}_{2} \mathrm{~S}, K_{\mathrm{al}}=1.4 \times 10^{-7}\) and \(K_{\mathrm{a} 2}=1.0 \times 10^{-14}\). (a) \(0.1 \mathrm{M}\) (b) \(2.8 \times 10^{-20} \mathrm{M}\) (c) \(2.8 \times 10^{-22} \mathrm{M}\) (d) \(1.4 \times 10^{-20} \mathrm{M}\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

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