   Chapter 15, Problem 68QAP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
1 views

# 68. Aluminum ion may be precipitated from aqueous solution by addition of hydroxide ion, forming Al(OH)3. A large excess of hydroxide ion must not be added, however, because the precipitate of Al(OH)3 will redissolve as a soluble compound containing aluminum ions and hydroxide ions begins to form. How many grains of solid NaOH should be added to 10.0 mL of 0.250 M A1Cl3 to just precipitate all the aluminum?

Interpretation Introduction

Interpretation:

The grams of solid NaOH that should be added to AlCl3 solution to precipitate all the aluminum is to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The moles of any element is calculated by the formula,

Moles=MassgMolarmass

Molarity is used to find out the concentration of solution.

The molarity is calculated by the formula,

Molarity=NumberofmolesofsoluteVolumeofsolutionL.

Explanation

The value of volume and molarity of AlCl3 is given to be 10.0mL and 0.250M respectively.

The conversion of units of 10.0mL into L is done as,

10.0mL=10.01000L=0.01L

The number of moles of AlCl3 solution is calculated by the formula,

Numberofmoles=Molarity×Volumeofsolution        (1)

Substitute the values of molarity and volume of AlCl3 solution in the equation (1).

Numberofmoles=0.250M×0.01L=0.0025moles

Sodium hydroxide reacts with aluminum chloride to form aluminum hydroxide. The balanced equation for sodium hydroxide and aluminum chloride is shown below.

3NaOH+AlCl3AlOH3+3NaCl

The above reaction indicates that one equivalent of AlCl3 reacts with three equivalents of sodium hydroxide

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