/*! 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 27 Calculate the solubility (g/100 ... [FREE SOLUTION] | 91影视

91影视

Calculate the solubility (g/100 mL) of iron(II) hydroxide in buffered solutions with the following pH's. (a) 4 (b) 7 (c) 10

Short Answer

Expert verified
The solubility of iron(II) hydroxide in buffered solutions with pH levels 4, 7, and 10 is 0.219 g/100 mL, 0.438 g/100 mL, and 4.38 脳 10鈦烩伔 g/100 mL, respectively.

Step by step solution

01

Write the balanced chemical equation for the dissolution of Fe(OH)鈧

The balanced chemical equation for the dissolution of iron(II) hydroxide is: Fe(OH)鈧(s) 鈬 Fe虏鈦(aq) + 2OH鈦(aq)
02

Find the solubility product constant (Ksp) of Fe(OH)鈧

The Ksp of Fe(OH)鈧 is 4.87 脳 10鈦宦光伔 (found in a chemistry reference table). This constant represents the equilibrium between the dissolved ions and the solid compound.
03

Write the equation for the relationship between pH and the concentration of hydroxide ions

The relationship between pH and the concentration of hydroxide ions is given by the following formula: \[pH + pOH = 14\] Since pOH = -log[OH鈦籡, we can rewrite the equation as: \[pH + -\log[OH鈦籡 = 14\]
04

Determine the concentration of hydroxide ions for each pH level

Use the relationship derived in step 3 to calculate the concentration of OH鈦 ions for each pH level: (a) For pH 4: \[4 + -\log[OH鈦籡 = 14\] \[10^{-10} = [OH鈦籡\] (b) For pH 7: \[7 + -\log[OH鈦籡 = 14\] \[10^{-7} = [OH鈦籡\] (c) For pH 10: \[10 + -\log[OH鈦籡 = 14\] \[10^{-4} = [OH鈦籡\]
05

Calculate the concentration of Fe虏鈦 ions using the Ksp expression

Using the Ksp expression, we can calculate the concentration of Fe虏鈦 ions for each pH level: Ksp = [Fe虏鈦篯[OH鈦籡虏 For each pH level: (a) For pH 4: \[4.87 脳 10^{-17} = [Fe虏鈦篯(10^{-10})^2\] \[2.44 脳 10^{-3} = [Fe虏鈦篯\] (b) For pH 7: \[4.87 脳 10^{-17} = [Fe虏鈦篯(10^{-7})^2\] \[4.87 脳 10^{-3} = [Fe虏鈦篯\] (c) For pH 10: \[4.87 脳 10^{-17} = [Fe虏鈦篯(10^{-4})^2\] \[4.87 脳 10^{-9} = [Fe虏鈦篯\]
06

Convert the concentration of Fe虏鈦 ions to grams per 100 mL

In order to find the solubility in grams per 100 mL, we will multiply the molar concentration of Fe虏鈦 by the molar mass of Fe(OH)鈧 and then multiply by 100 mL: Solubility = [Fe虏鈦篯 脳 (molar mass of Fe(OH)鈧) 脳 100 mL For each pH level: (a) For pH 4: Solubility = (2.44 脳 10鈦宦 mol/L) 脳 (89.86 g/mol) 脳 0.1 L = 0.219 g (b) For pH 7: Solubility = (4.87 脳 10鈦宦 mol/L) 脳 (89.86 g/mol) 脳 0.1 L = 0.438 g (c) For pH 10: Solubility = (4.87 脳 10鈦烩伖 mol/L) 脳 (89.86 g/mol) 脳 0.1 L = 4.38 脳 10鈦烩伔 g The solubility of iron(II) hydroxide in buffered solutions with pH levels 4, 7, and 10 is 0.219 g/100 mL, 0.438 g/100 mL, and 4.38 脳 10鈦烩伔 g/100 mL, respectively.

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.

iron(II) hydroxide
Iron(II) hydroxide, chemically represented as Fe(OH) _{2}, is an insoluble solid under typical conditions. When it dissolves in water, it dissociates into iron ions (Fe虏鈦) and hydroxide ions (OH鈦). The dissolution of iron(II) hydroxide involves the balanced chemical reaction:

Fe(OH)鈧(s) 鈬 Fe虏鈦(aq) + 2OH鈦(aq)

Understanding this equilibrium is crucial in determining how much of the compound will actually dissolve. This compound is known for its role in various chemical processes, such as water treatment and certain redox reactions. However, due to its low solubility, it is relatively inert in neutral or weakly acidic conditions.
solubility product constant (Ksp)
The solubility product constant, or Ksp, is a measure of the solubility of sparingly soluble compounds. It represents the product of the molar concentrations of the ions in a saturated solution, each raised to the power of their coefficients in the balanced equation.

For iron(II) hydroxide, the Ksp expression is:

\[ K_{sp} = [Fe^{2+}][OH^-]^2 \]

In this exercise, the Ksp is given as 4.87 脳 10鈦宦光伔, indicating a very low solubility. By manipulating the Ksp, you can determine the concentrations of the individual ions in the solution. Since Ksp is a constant at a given temperature, it allows chemists to predict the extent to which a compound will dissolve. This is especially useful in applications like predicting the amount of a substance that can be dissolved at a particular pH level.
pH and pOH relationship
The pH and pOH are interconnected aspects of a solution's acidity or basicity. They follow the relationship:

\[ pH + pOH = 14 \]

This equation is pivotal for converting between pH, which measures the acidity, and pOH, which measures the basicity. The hydroxide ion concentration is linked to pOH, while the hydrogen ion concentration is linked to pH. For instance, if the pH of a solution is known to be 7, it means the pOH is also 7, indicating a neutral solution. This balance is critical for maintaining different chemical environments and is exploited in buffer solutions to resist drastic pH changes. It helps chemists regulate reactions that are sensitive to pH fluctuations.
hydroxide ion concentration
Hydroxide ion concentration ([OH鈦籡) plays a crucial role in determining the solubility of compounds like iron(II) hydroxide. Using the relationship between pH and pOH, you can calculate the concentration of hydroxide ions. The equation to find [OH鈦籡 from pH is:

\[ pOH = 14 - pH \]
\[ [OH^-] = 10^{-pOH} \]

A higher [OH鈦籡 results in greater solubility for negatively charged ions by shifting the equilibrium of a dissolution reaction. In our iron(II) hydroxide example, as the pH increases (which decreases pOH), the hydroxide concentration also increases, altering the solubility of the compound. This is why solubility changes significantly with pH adjustments. Understanding hydroxide ion concentration helps with practical applications, including formulating buffered solutions to maintain desired reaction conditions.

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 made up by mixing \(125 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{AuNO}_{3}\) and \(225 \mathrm{~mL}\) of \(0.049 \mathrm{M} \mathrm{AgNO}_{3}\). Twenty-five mL of a \(0.0100 \mathrm{M}\) solution of \(\mathrm{HCl}\) is then added. \(K_{\text {sp }}\) of \(\mathrm{AuCl}=2.0 \times 10^{-13} .\) When equilibrium is established, will there be- -no precipitate? -a precipitate of \(\mathrm{AuCl}\) only? -a precipitate of \(\mathrm{AgCl}\) only? -a precipitate of both \(\mathrm{AgCl}\) and \(\mathrm{AuCl}\) ?

Consider the insoluble salts \(\mathrm{JQ} \mathrm{K}_{2} \mathrm{R}, \mathrm{L}_{2} \mathrm{~S}_{3}, \mathrm{MT}_{2}\), and \(\mathrm{NU}_{3}\). They are formed from the metal ions \(\mathrm{J}^{+}, \mathrm{K}^{+}, \mathrm{L}^{3+}, \mathrm{M}^{2+}\), and \(\mathrm{N}^{3+}\) and the nonmetal ions \(\mathrm{Q}^{-}, \mathrm{R}^{2-}, \mathrm{S}^{2-}, \mathrm{T}^{-}\), and \(\mathrm{U}^{-}\). All the salts have the same \(K_{\text {sp }}, 1 \times 10^{-10}\), at \(25^{\circ} \mathrm{C}\). (a) Which salt has the highest molar solubility? (b) Does the salt with the highest molar solubility have the highest solubility in g salt/100 g water? (c) Can the solubility of each salt in \(\mathrm{g} / 100 \mathrm{~g}\) water be determined from the information given? If yes, calculate the solubility of each salt in \(\mathrm{g} / 100 \mathrm{~g}\) water. If no, why not?

Write net ionic equations for the reactions of each of the following with strong acid. (a) \(\mathrm{CaCO}_{3}\) (b) NiS (c) \(\mathrm{Al}(\mathrm{OH})_{3}\) (d) \(\mathrm{Sb}(\mathrm{OH})_{4}^{-}\) (e) \(\mathrm{AgCl}\)

Solid lead nitrate is added to a solution that is \(0.020 \mathrm{M}\) in \(\mathrm{OH}^{-}\) and \(\mathrm{SO}_{4}^{2-}\). Addition of the lead nitrate does not change the volume of the solution. (a) Which compound, \(\mathrm{PbSO}_{4}\) or \(\mathrm{Pb}(\mathrm{OH})_{2}\left(K_{s p}=2.8 \times 10^{-16}\right)\), will precipitate first? (b) What is the \(\mathrm{pH}\) of the solution when \(\mathrm{PbSO}_{4}\) first starts to precipitate?

Write a net ionic equation for the reaction with \(\mathrm{OH}^{-}\) by which (a) \(\mathrm{Sb}^{3+}\) forms a precipitate. (b) antimony(III) hydroxide dissolves when more \(\mathrm{OH}^{-}\) is added. (c) \(\mathrm{Sb}^{3+}\) forms a complex ion.

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.