   Chapter 15, Problem 63QAP ### 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
17 views

# 63. The amount of nickel(II) present in an aqueous solution can be determined by precipitating the nickel with the organic chemic al reagent dimethylglyoxime [CH3C(NOH)C(NOH)CH3, commonly abbreviated as “DMG”]. Ni 2+ ( a q ) + 2DMG ( aq ) → Ni ( DMG ) 2 ( s ) How many milliliters of 0.0703 M DMG solution is required to precipitate all the nickel(Il) present in 10.0 mL of 0.103 M nickel(II) sulfate solution?

Interpretation Introduction

Interpretation:

The milliliters of 0.0703M

DMG solution which is required to precipitate all the nickel(II) present in a given solution 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 number of moles 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 molarity and volume of nickel(II) sulfate solution is given to be 0.103M and 10.0mL respectively.

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

10.0mL=10.01000L=0.01L

The number of moles of Ni2+ is calculated by the formula,

NumberofmolesofNi2+=Molarity×VolumeofsolutionL        (1)

Substitute the values of molarity and volume of solution containing Ni2+ in the equation (1).

NumberofmolesofNi2+=0.103M×0.01L=0.00103moles

The given chemical equation is shown below.

Ni2+aq+2DMGaqNiDMG2s

The given equation exhibits that one equivalent of Ni2+ and two equivalents of DMG combine to form nickel(II) sulfate solution

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