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Introductory Chemistry: A Foundati...

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

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BuyFindarrow_forward

Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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. Mercuric sulphide, HgS, is one of the least soluble salts known, with K sp = 1.6 × 10 54 at 25 °C. Calculate the solubility of HgS in moles per liter and in grams per liter.

Interpretation Introduction

Interpretation:

The solubility of mercury sulfide (HgS) in moles per liter and grams per liter is to be calculated.

Concept Introduction:

The equilibrium expression is used to calculate the solubility and solubility product constant of salts. The equilibrium expression for HgS is,

Ksp=[Hg2+][S2].

Explanation

The Ksp of HgS at 25°C is 1.6×1054.

It is assumed that x moles per liter of HgS, is dissolved to reach the equilibrium stage. Thus,

xmolLHgS(s)xmolLHg2+(aq)+xmolLS2(aq)

At equilibrium,

[Hg2+]=xmolL[S2]=xmolL

The equilibrium expression for HgS is,

Ksp=[Hg2+][S2]

Where,

  • [Hg2+] is the molar concentration of Hg2+.
  • [S2] is the molar concentration of S2.
  • Ksp is the solubility product constant of HgS.

Substitute the values of [Hg2+], [S2], and Ksp in above equation

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