/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Calculate the solubility of \(\m... [FREE SOLUTION] | 91影视

91影视

Calculate the solubility of \(\mathrm{Mn}(\mathrm{OH})_{2}\) in grams per liter when buffered at \(\mathrm{pH}\) (a) \(7.0,(\mathbf{b}) 9.5,(\mathbf{c}) 11.8\).

Short Answer

Expert verified
Solubility of Mn(OH)鈧 increases with higher pH values: lower at pH 7, higher at pH 11.8 due to [OH鈦籡 concentration.

Step by step solution

01

Write the Dissolution Equation

The dissolution of manganese(II) hydroxide in water can be represented by the equilibrium equation: \[ \text{Mn(OH)}_2 (s) \rightleftharpoons \text{Mn}^{2+} (aq) + 2 \text{OH}^- (aq) \].
02

Express the Solubility Product, Ksp

The solubility product (\( K_{sp} \)) for Mn(OH)鈧 can be expressed as: \[ K_{sp} = [\text{Mn}^{2+}][\text{OH}^-]^2 \]. Assume the solubility of Mn(OH)鈧 is \( s \), then \([\text{Mn}^{2+}] = s \) and \([\text{OH}^-] = 2s \).
03

pH and pOH Relationship

We know that \( \text{pH} + \text{pOH} = 14 \). Calculate \( \text{pOH} \) from the given \( \text{pH} \) values to find the concentration of \( \text{OH}^- \) ions, \([\text{OH}^-] = 10^{-\text{pOH}} \).
04

Solubility Adjustment for pH 7.0

At \( \text{pH} = 7.0 \), \( \text{pOH} = 14 - 7 = 7 \). Therefore, \([\text{OH}^-] = 10^{-7} \). Using \( K_{sp} = s \times (10^{-7})^2 \), calculate \( s \).
05

Solubility Adjustment for pH 9.5

For \( \text{pH} = 9.5 \), \( \text{pOH} = 14 - 9.5 = 4.5 \). Thus, \([\text{OH}^-] = 10^{-4.5} \). Substitute into the equation: \( K_{sp} = s \times (10^{-4.5})^2 \), and solve for \( s \).
06

Solubility Adjustment for pH 11.8

With \( \text{pH} = 11.8 \), \( \text{pOH} = 14 - 11.8 = 2.2 \). So, \([\text{OH}^-] = 10^{-2.2} \). Use the equation: \( K_{sp} = s \times (10^{-2.2})^2 \), and calculate \( s \).
07

Conversion to Grams per Liter

Convert the solubilities from molarity (mol/L) to grams/L by multiplying \( s \) by the molar mass of \( \text{Mn(OH)}_2 \). The molar mass is \( 54.94 + 2(16.00 + 1.01) = 88.95 \) g/mol. The solubility in grams per liter is \( s \times 88.95 \).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product Constant (Ksp)
The solubility product constant, abbreviated as Ksp, is a value that represents the solubility of a sparingly soluble compound in water. It quantifies the extent to which a solid will dissolve in water to form a saturated solution.
In simple terms, Ksp is used to tell us how much of the solid will dissolve. A higher Ksp value indicates greater solubility, while a lower Ksp value signifies limited solubility.
For any given compound like manganese(II) hydroxide (Mn(OH)8), we write a dissolution equation to represent it breaking down into its ions:
  • Mn(OH)8(s) 1 Mn2+ (aq) + 2 OH7 (aq)
The Ksp expression is then formulated based on these ions. For Mn(OH)8, the expression is:
  • \[ K_{sp} = [\text{Mn}^{2+}][\text{OH}^-]^2 \]
Here, the concentration of OH^- is squared because there are two OH^- ions for each formula unit of Mn(OH)8. Knowing the Ksp allows us to calculate how the solubility adjusts under different conditions, like changes in pH, which are fundamental in solubility calculations.
Acid-Base Chemistry
Acid-base chemistry plays a significant role in understanding solubility, especially when dealing with substances that produce or consume H+ and OH- ions. The pH of a solution is a measure of its acidity or basicity.
The pH scale ranges from 0 to 14, where:
  • A pH less than 7 indicates an acidic solution.
  • A pH greater than 7 means the solution is basic.
  • A pH of 7 is neutral.
The relationship between pH and pOH is expressed as:
  • \[ \text{pH} + \text{pOH} = 14 \]
This relationship helps us calculate the concentration of hydroxide ions [OH^-], especially in solutions where an acid or base affects solubility.
The concentration of [OH^-] is determined as \(10^{-\text{pOH}}\). As pH changes, it alters the concentrations of the ions in the solution, subsequently affecting the solubility of compounds like Mn(OH)8. Thus, understanding acid-base chemistry is crucial in adjusting solubility for different pH levels.
Equilibrium Reactions
Equilibrium reactions occur when a reversible chemical reaction reaches a state where the forward and reverse reaction rates are equal. This means no net change in the concentrations of reactants and products.
In the context of solubility, equilibrium reactions describe the balance between a solid and its dissolved ions in solution:
  • Mn(OH)8(s) \(\rightleftharpoons\) Mn2+ (aq) + 2 OH7 (aq)
At equilibrium, the equilibrium expression, or Ksp in this case, quantifies the product of the concentrations of the ions raised to their stoichiometric coefficients.
This helps in understanding how the balance shifts when conditions such as concentration, temperature, or pH change. If external conditions change, Le Chatelier's principle predicts how the equilibrium will shift to accommodate this change, often impacting solubility.
Evaluating equilibrium reactions in solubility is essential for calculating exact solubility values, particularly when additional factors, like common ions, can shift the equilibrium.

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

Predict whether the equivalence point of each of the following titrations is below, above, or at pH 7: (a) \(\mathrm{NaHCO}_{3}\) titrated with \(\mathrm{NaOH},(\mathbf{b}) \mathrm{NH}_{3}\) titrated with \(\mathrm{HCl},(\mathbf{c}) \mathrm{KOH}\) titrated with \(\mathrm{HBr}\).

In nonaqueous solvents, it is possible to react HF to create \(\mathrm{H}_{2} \mathrm{~F}^{+} .\) Which of these statements follows from this observation? (a) HF can act like a strong acid in nonaqueous solvents, (b) HF can act like a base in nonaqueous solvents, (c) HF is thermodynamically unstable, \((\mathbf{d})\) There is an acid in the nonaqueous medium that is a stronger acid than HE.

The following boxes represent aqueous solutions containing a weak acid, HA and its conjugate base, \(A^{-}\). Water molecules, hydronium ions, and cations are not shown. Which solution has the highest pH? Explain. [Section 17.1]

Consider a beaker containing a saturated solution of CaF \(_{2}\) in equilibrium with undissolved \(\mathrm{CaF}_{2}(s)\). Solid \(\mathrm{CaCl}_{2}\) is then added to the solution. (a) Will the amount of solid \(\mathrm{CaF}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Ca}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of F ions in solution increase or decrease?

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) and whose corresponding \(K_{\text {ip }}=6.8 \times 10^{-27}\). As discussed in the Chemistry and Life box on page 790 , fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F},\) whose \(K_{y p}=1.0 \times 10^{-60}\) (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

See all solutions

Recommended explanations on Chemistry Textbooks

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.