   Chapter 15, Problem 95AP ### 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

# 95. Many metal ions form insoluble sulfide compounds when a solution of the metal ion is treated with hydrogen sulfide gas. For example, nickel(II) precipitates nearly quantitatively as NiS when H2S gas is bubbled through a nickel ion solution. How many milliliters of gaseous H2S at STP are needed to precipitate all (he nickel ion present in 10. mL of 0.050 M NiCl2 solution?

Interpretation Introduction

Interpretation:

The milliliters of gaseous H2S at STP required to precipitate all the nickel ions present in a given solution is to be calculated.

Concept Introduction:

Solution is composed of solute and solvent particles. Solute particles are always present in lower amount as compared to amount of the solvent in the solution. 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 molarity is calculated by the formula,

Molarity=NumberofmolesofsoluteVolumeofsolutionL.

Explanation

The molarity and volume of NiCl2 solution is given to be 0.050M and 10.mL respectively.

The conversion of units of 10.mL into L is done as,

10.mL=10.1000L=0.01L

The number of moles of nickel ions in a solution is calculated by the formula,

Moles=Molarity×Volumeofsolution        (1)

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

Moles=0.050M×0.01L=0.0005moles

The equation when NiCl2 reacts with H2S is shown below.

NiCl2aq+H2SaqNiS+2HClaq

The above equation implies that one equivalent of NiCl2 combines with one equivalent of H2S to form precipitates of NiS. Thus, the same number of moles of H2S required to precipitate nickel ions

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