Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 105

What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{S}\) and maintained at \(\mathrm{pH} 3.0\) with HCl?What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10\) \(M\) \(\mathrm{H}_{2} \mathrm{S}\) and maintained at \(\mathrm{pH} 3.0\) with \(\mathrm{HCl}\)?

Short Answer

Expert verified
The maximum possible concentration of Ni²⁺ ion in water at 25°C that is saturated with 0.10 M H₂S and maintained at pH 3.0 with HCl is 3.3 x 10⁻⁴ M.

Step by step solution

01

Finding the [H⁺] concentration *at equilibrium*

Given that the pH of the solution is 3.0, we can calculate the [H⁺] concentration using the pH formula: pH = -log10[H⁺] 3.0 = -log10[H⁺] Solving for [H⁺]: [H⁺] = 10⁻³M
02

Calculating the [HS⁻] concentration at equilibrium

Using the Ka1 value and the equilibrium expression for H₂S: Ka1 = [H⁺][HS⁻] / [H₂S] Rearranging to solve for [HS⁻]: [HS⁻] = (Ka1 * [H₂S]) / [H⁺] Plugging in the given values for [H₂S] = 0.10M and [H⁺] = 10⁻³M: [HS⁻] = (9.1 x 10⁻⁸ * 0.10) / 10⁻³ [HS⁻] = 9.1 x 10⁻⁶M
03

Calculating the [S²⁻] concentration at equilibrium

Using the Ka2 value and the equilibrium expression for HS⁻: Ka2 = [H⁺][S²⁻] / [HS⁻] Rearranging to solve for [S²⁻]: [S²⁻] = (Ka2 * [HS⁻]) / [H⁺] Plugging in the calculated [HS⁻] = 9.1 x 10⁻⁶M and [H⁺] = 10⁻³M: [S²⁻] = (1.0 x 10⁻¹³ * 9.1 x 10⁻⁶) / 10⁻³ [S²⁻] = 9.1 x 10⁻¹⁰M
04

Finding the maximum possible concentration of Ni²⁺

Now, we use the Ksp value and the equilibrium expression for NiS: Ksp = [Ni²⁺][S²⁻] Rearranging to solve for [Ni²⁺]: [Ni²⁺] = Ksp / [S²⁻] Plugging in the calculated [S²⁻] = 9.1 x 10⁻¹⁰M and Ksp = 3.0 x 10⁻¹³: [Ni²⁺] = (3.0 x 10⁻¹³) / (9.1 x 10⁻¹⁰) [Ni²⁺] = 3.3 x 10⁻⁴M So, the maximum possible concentration of Ni²⁺ ion in water at 25°C that is saturated with 0.10 M H₂S and maintained at pH 3.0 with HCl is 3.3 x 10⁻⁴ M.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A solution is \(1 \times 10^{-4} M\) in \(\mathrm{NaF}, \mathrm{Na}_{2} \mathrm{S},\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\mathrm{sp}}\) values are \(K_{\mathrm{sp}}\left(\mathrm{PbF}_{2}\right)\) \(=4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29},\) and \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=\) \(1 \times 10^{-54}\).

Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}\) b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate (EDTA \(^{4-}\) ) is used as a complexing agent in chemical analysis and has the following structure: Solutions of EDTA \(^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of EDTA \(^{4-}\) with \(\mathrm{Pb}^{2+}\) is $$\begin{aligned} \mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) & \\ K &=1.1 \times 10^{18} \end{aligned}$$ Consider a solution with 0.010 mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to \(1.0 \mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050 \(M\) \(\mathrm{Na}_{4} \mathrm{EDTA}\). Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

Calculate the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=5.9 \times 10^{-11}\).

The \(K_{\mathrm{sp}}\) for silver sulfate \(\left(\mathrm{Ag}_{2} \mathrm{SO}_{4}\right)\) is \(1.2 \times 10^{-5} .\) Calculate the solubility of silver sulfate in each of the following. a. water b. \(0.10 M\) AgNO \(_{3}\) c. \(0.20 M \mathrm{K}_{2} \mathrm{SO}_{4}\)

Which of the following will affect the total amount of solute that can dissolve in a given amount of solvent? a. The solution is stirred. b. The solute is ground to fine particles before dissolving. c. The temperature changes.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free