   Chapter 16, Problem 112IP

Chapter
Section
Textbook Problem

# The Ksp for Q, a slightly soluble ionic compound composed of M22+ and X− ions, is 4.5 × 10−29. The electron configuration of M+ is [Xe]6s14f145d10. The X− anion has 54 electrons. What is the molar solubility of Q in a solution of NaX prepared by dissolving 1.98 g NaX in 150. mL solution?

Interpretation Introduction

Interpretation: The solubility product of a slightly soluble compound is given. Electronic configuration of its cation and number of electrons in the anion is given. The molar solubility of the compound in a solution of NaX   prepared by dissolving 1.98g of NaX in 150mL  solution is to be calculated.

Concept introduction: The solubility product is the mathematical product of a substance’s dissolved ion concentration raised to its power of its stoichiometric coefficients. When sparingly soluble ionic compound releases ions in the solution, it gives relevant solubility product. The solvent is generally water.

Explanation

Explanation

To determine: The molar solubility of the given compound.

The slightly soluble ionic compound is Hg2I2 and the compound NaX is NaI .

The given electronic configuration of cation M+ is [Xe]6s14f145d10 .

Number of electrons in xenon is 54. Thus the total number of electrons of cation M+ is 79 electrons. Sine M+ is formed when one electron is removed from element M , the number of electrons in element M is 80 .

According to periodic table, the element having 80 electrons is mercury (Hg) .

Number of electrons of anion X is 54. It is formed when one electron is removed from element X. The element having 53 electrons is iodine (I) .

Therefore, the slightly soluble ionic compound is Hg2I2 and the compound NaX is NaI .

The molarity of NaI is 0.0881 mol/L_ .

Given

The given mass of NaI is 1.98g .

Volume of the solution is 150mL .

The molar mass of NaI=149.89 g/mol

The number of moles can be calculated by the formula,

Moles=GivenmassMolarmass

Substitute the values of molar mass and given mass in above equation.

Moles=1.98149.89

The conversion of mL to L is done as,

1mL=103L

Therefore, the conversion of 150mL to L is done as,

150mL=150×103L

Molarity of NaI is number of moles of NaI in 1L of solution.

Therefore, molarity of NaI =1.98149.89×150×103=0.0881 mol/L_

The molarity of NaI is 0.0881 mol/L_ .

The molar solubility of Q(Hg2I2) is 5.8×10-27mol/L_

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