/*! 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 42 Calculate the solubility of \(\m... [FREE SOLUTION] | 91影视

91影视

Calculate the solubility of \(\mathrm{Co}(\mathrm{OH})_{2}(s)\left(K_{\mathrm{sp}}=2.5 \times 10^{-16}\right)\) in a buffered solution with a \(\mathrm{pH}\) of 11.00.

Short Answer

Expert verified
The solubility of Co(OH)鈧 in a buffered solution with a pH of 11.00 is calculated using the given Ksp value and the concentration of OH鈦 ions. Using the dissolution of Co(OH)鈧 in water, the balanced chemical equation, and the Ksp expression, we find the concentration of Co虏鈦 ions. After converting the given pH to the concentration of OH鈦 ions and substituting the given Ksp and [OH鈦籡 into the Ksp expression, we solve for the concentration of Co虏鈦 ions. The solubility of Co(OH)鈧 in the buffered solution is found to be 2.5 脳 10鈦宦光伆 M.

Step by step solution

01

Write the balanced chemical equation

The dissolution of Co(OH)2 in water can be represented by the following balanced chemical equation: Co(OH)2 (s) 鈫 Co2+ (aq) + 2 OH- (aq)
02

Write the expression for Ksp

The solubility product constant (Ksp) for the dissolution of Co(OH)2 is given by: Ksp = [Co虏鈦篯 脳 [OH鈦籡虏 where [Co虏鈦篯 and [OH鈦籡 are the equilibrium concentrations of Co2+ and OH- ions, respectively. We are given that Ksp = 2.5 脳 10鈦宦光伓.
03

Convert the given pH to [OH鈦籡 concentration

We are given the pH of the buffered solution as 11.00. To find the concentration of OH- ions, we will first find the pOH: pOH = 14 - pH = 14 - 11 = 3 Now, we can calculate the concentration of OH- ions using the pOH: [OH鈦籡 = 10^(-pOH) = 10鈦宦 M
04

Substitute the given Ksp and [OH鈦籡 into the Ksp expression

Now that we have the concentration of OH- ions, we can substitute it into the Ksp expression: 2.5 脳 10鈦宦光伓 = [Co虏鈦篯 脳 (10鈦宦)虏 2.5 脳 10鈦宦光伓 = [Co虏鈦篯 脳 10鈦烩伓
05

Solve for the concentration of Co虏鈦 ions

Now we will solve for the equilibrium concentration of Co虏鈦 ions: [Co虏鈦篯 = (2.5 脳 10鈦宦光伓) / 10鈦烩伓 [Co虏鈦篯 = 2.5 脳 10鈦宦光伆 M
06

Calculate the solubility of Co(OH)2

The solubility of Co(OH)2 in the given buffered solution is equal to the equilibrium concentration of Co虏鈦 ions: Solubility of Co(OH)2 = 2.5 脳 10鈦宦光伆 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.

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.

Ksp
When discussing solubility, the solubility product constant, or Ksp, is a key concept. It's a measure of how much of a compound can dissolve in a solution before it begins to precipitate. Imagine Ksp as a balance between the solid and its ions in solution. For cobalt(II) hydroxide, the equation is expressed as:\[ \text{Co(OH)}_2 (s) \rightleftharpoons \text{Co}^{2+} (aq) + 2\text{OH}^{-} (aq) \]The Ksp equation becomes:\[ K_{sp} = [\text{Co}^{2+}] \times [\text{OH}^-]^2 \]This helps us find the equilibrium concentrations of ions, acting as a snapshot of how much dissolves before reaching saturation. It's fascinating how small changes in the solution can drastically shift the equilibrium.
Chemical Equilibrium
Chemical equilibrium is a dynamic state where the concentration of reactants and products remains constant over time. For solubility, equilibrium means the rate at which Co(OH)鈧 dissolves is equal to the rate at which it crystallizes. Here's how it works:
  • The system balances between the dissolved ions and the undissolved solid.
  • Even at equilibrium, reactions do not stop; they just proceed at equal rates.
  • The concentrations of ions are governed by the Ksp value, setting the maximum solubility under given conditions.
This dynamic process allows predictions about solubility and guides us in tweaking conditions to change how much can dissolve.
pH and pOH
Understanding pH and pOH is essential when dealing with buffered solutions, as these influence solubility. pH represents the acidity, and pOH the alkalinity, both summing to 14 in a solution:\[ \text{pH} + \text{pOH} = 14 \]For the solution with a pH of 11, we find the pOH is 3. This reveals important details:
  • A high pH (or low pOH) indicates a basic environment.
  • Such conditions shift equilibrium, affecting how many ions the Co(OH)鈧 can release.
  • In this instance, more OH鈦 ions are already present, impacting how much cobalt can dissolve.
It's like adjusting a knob that influences the entire solution's behavior, reflecting the interplay of hydrogen and hydroxide ions.
Buffer Solutions
Buffer solutions maintain a stable pH, crucial for precise solubility calculations. They are mixtures of weak acids and their conjugate bases, or vice versa. Here's what they achieve:
  • Resist changes in pH, even when strong acids or bases are added.
  • Provide a controlled environment to study reactions like Co(OH)鈧 solubility.
  • Allow us to assume certain concentrations, like the OH鈦 found previously, as constant.
Buffers act like a pH shock absorber, making them indispensable in experiments where precise pH levels are vital. This control ensures solubility is mainly influenced by intrinsic properties like Ksp, rather than fluctuating pH levels, offering a reliable window into the compound's behavior.

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 prepared by mixing 100.0 \(\mathrm{mL}\) of \(1.0 \times 10^{-2} \mathrm{M}\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and 100.0 \(\mathrm{mL}\) of \(1.0 \times 10^{-3} \mathrm{M} \mathrm{NaF} .\) Will \(\mathrm{PbF}_{2}(s)\) \(\left(K_{\mathrm{sp}}=4 \times 10^{-8}\right)\) precipitate?

A solution is prepared by adding 0.10 mole of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to 0.50 \(\mathrm{L}\) of \(3.0 M\) \(\mathrm{NH}_{3}\) . Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text { overall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{8}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ for the overall reaction $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$

Nanotechnology has become an important field, with applications ranging from high-density data storage to the design of 鈥渘ano machines.鈥 One common building block of nano structured architectures is manganese oxide nano particles. The particles can be formed from manganese oxalate nano rods, the formation of which can be described as follows: $$\mathrm{Mn}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MnC}, \mathrm{O}_{4}(a q) \quad K_{1}=7.9 \times 10^{3}$$ $$\mathrm{MnC}_{2} \mathrm{O}_{4}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}(a q) \quad K_{2}=7.9 \times 10^{1}$$ Calculate the value for the overall formation constant for \(\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-} :\) $$K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2}}$$

Assuming that the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is \(1.6 \times 10^{-7} \mathrm{mol} / \mathrm{L}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{sp}}\) for this salt. Ignore any potential reactions of the ions with water.

Write equations for the step wise formation of each of the following complex ions. a. \(N i(C N)_{4}^{2-}\) b. \(V\left(C_{2} O_{4}\right)_{3}^{3-}\)

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.