   Chapter 16, Problem 49E

Chapter
Section
Textbook Problem

# Will a precipitate form when 100.0 mL of 4.0 × 10−4 M Mg(NO3)2 is added to 100.0 mL of 2.0 × 10−4 M NaOH?

Interpretation Introduction

Interpretation: The concentration and volume of Mg(NO3)2 and NaOH is given. The formation of a precipitate, when the given amount of Mg(NO3)2 is added to NaOH solution is to be validated.

Concept introduction: The formation of a solid in a solution is known as precipitation.

Solubility product, Ksp is defined as the concentration of ions in a saturated solution where each ion is raised to the power of their coefficients. Ion product, Q is defined as the initial concentration of ions in any solution where each ion is raised to the power of their coefficients.

The relation between ion product, Q and solubility product, Ksp is as follows.

If Ksp>Q , then no precipitate will form.

If Ksp<Q , then precipitate will form.

If Ksp=Q , then no precipitate will form.

Explanation

Explanation

To determine: If the formation of a precipitate takes place when the given amount of Mg(NO3)2 is added to NaOH .

The solubility product of Mg(OH)2 is 1.6×1011_ .

Given

Concentration of Mg(NO3)2 is 4.0×104M .

Concentration of NaOH is 2.0×104M .

The Mg(OH)2 is the only precipitate formed during mixing of 100mL NaOH and 100mL of Mg(NO3)2 if,

Ksp<Q

The reaction of Mg(NO3)2 and Mg(NO3)2 is,

Mg(NO3)2+NaOHMg(OH)2+NaNO3

Formula

The solubility product of Mg(OH)2 is calculated as,

Ksp=[Mg2+][OH]2

Where,

Ksp is the solubility product.

[Mg2+] is the concentration of Mg2+ .

[OH] is the concentration of OH .

Substitute the value of [OH] and [Mg2+] in the above expression.

Ksp=[Mg2+][OH]2=(4.0×104)(2.0×104)2=1.6×1011_

The initial concentration of Mg2+ is 2.0×104M_ and initial concentration of OH is 1.0×104M_ .

Given

Concentration of Mg(NO3)2 is 4.0×104M .

Concentration of NaOH is 2.0×104M .

Volume of NaOH is 100mL .

Volume of Mg(NO3)2 is 100mL .

It is assumed that the concentration of Mg2+ and OH is same as the concentration of Mg(NO3)2 and NaOH . Hence, concentration of Mg2+ and OH is,

[Mg2+]=4.0×104M[OH]=2.0×104M

Formula

The initial concentration of compound is calculated using the formula,

Substitute the values of volume, given concentration of Mg2+ and volume of original solution in the above equation

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