The solubility product of \(\mathrm{Co}(\mathrm{OH})_{3}\) is \(2.7 \times 10^{-43}\). The pH of saturated solution of \(\mathrm{Co}(\mathrm{OH})_{3}\) is about (a) \(7.0\) (b) \(11.0\) (c) \(3.0\) (d) \(3.48\)

Short Answer

Expert verified
The pH of saturated solution of \(\mathrm{Co}(\mathrm{OH})_{3}\) is about 3.48, so the answer is (d).

Step by step solution

01

Determine the balanced equation for dissolution

The first step is to understand the dissolution reaction of \(\mathrm{Co}(\mathrm{OH})_{3}\) in water. It dissolves to form \(\mathrm{Co}^{3+}\) and \(\mathrm{OH}^{-}\) ions in the following manner: \[ \mathrm{Co}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Co}^{3+}(aq) + 3\mathrm{OH}^{-}(aq) \]
02

Write the expression for the solubility product constant

The solubility product constant expression, \(K_{sp}\), for \(\mathrm{Co}(\mathrm{OH})_{3}\) can be written as: \[ K_{sp} = [\mathrm{Co}^{3+}][\mathrm{OH}^{-}]^3 \] where the square brackets represent the molar concentrations of the ions at equilibrium.
03

Express the concentrations in terms of solubility

Let \(s\) be the solubility of \(\mathrm{Co}(\mathrm{OH})_{3}\), which is the molarity of \(\mathrm{Co}^{3+}\) in a saturated solution. The concentration of \(\mathrm{OH}^{-}\) will be \(3s\) because one \(\mathrm{Co}(\mathrm{OH})_{3}\) molecule produces three hydroxide ions. Thus, the expression can be rewritten as: \[ K_{sp} = s(3s)^3 = 27s^4 \]
04

Solve for the solubility \(s\)

Plugging the given \(K_{sp}\) value into the above expression, we get: \[ 2.7 \times 10^{-43} = 27s^4 \Rightarrow s^4 = \frac{2.7 \times 10^{-43}}{27} \Rightarrow s^4 = 10^{-44} \Rightarrow s = 10^{-11} \] This is the molarity of \(\mathrm{Co}^{3+}\) and also the solubility of \(\mathrm{Co}(\mathrm{OH})_{3}\) in mol/L.
05

Calculate the concentration of \(\mathrm{OH}^{-}\) ions

The concentration of \(\mathrm{OH}^{-}\) ions is three times the solubility of \(\mathrm{Co}(\mathrm{OH})_{3}\): \[ [\mathrm{OH}^{-}] = 3s = 3 \times 10^{-11} \]
06

Calculate the \(pOH\) of the solution

Using the \(pOH\) definition, we find: \[ pOH = -\log [\mathrm{OH}^{-}] = -\log(3 \times 10^{-11}) \approx 10.52 \] (using a calculator to find the log value)
07

Convert \(pOH\) to \(pH\)

The relationship between \(pH\) and \(pOH\) is given by \( pH + pOH = 14 \), for water at 25°C. Therefore, the \(pH\) of the solution is: \[ pH = 14 - pOH = 14 - 10.52 = 3.48 \]

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Key Concepts

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

Dissolution Reaction
When a solid compound, like Co(OH)3, dissolves in water, it undergoes a process called the dissolution reaction. This type of reaction is fundamental in understanding solubility and involves the breaking apart of the compound into its constituent ions. In our case, the Co(OH)3 dissociates into one Co3+ ion and three OH- ions, forming an equilibrium between the solid phase and its dissolved ions in solution.

The dissolution reaction can be influenced by various factors like temperature, the presence of other ions in the solution, and the common ion effect, which occurs when one of the ions produced by the dissolution is already present in the solution. Understanding the dissolution reaction is crucial for predicting solubility and for calculating the concentrations of ions in a solution, which has many practical applications, including those in environmental science, medicine, and engineering.
pH Calculation
The pH calculation is a critical aspect when dealing with solutions in chemistry, especially in the context of solubility products. pH is a measure of the acidity or basicity of an aqueous solution, and it is directly related to the concentration of hydrogen ions (H+) in the solution. The pH of a solution is calculated as the negative logarithm (base 10) of the H+ ion concentration, typically expressed in moles per liter (M).

In our exercise, we don't directly calculate the pH from H+, but instead, we focus on the hydroxide ions (OH-) concentration to first find the pOH, which is similarly defined as the negative logarithm of the OH- ion concentration. The pH is then found by subtracting the pOH from 14, which is based on the relationship between pH and pOH at room temperature. Knowing how to correctly perform pH calculations is essential for scientists and students, impacting other areas of study like titrations, buffer solutions, and the biological relevance of pH in body fluids.
Equilibrium Concentration
Equilibrium concentration refers to the concentrations of reactants and products in a reaction mixture at the state of equilibrium. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, leading to constant concentrations of all species involved. In our exercise with Co(OH)3, the equilibrium concentrations of Co3+ and OH- are directly tied to the solubility product constant, Ksp.

The concept of equilibrium concentration is not limited to solubility reactions; it's also applicable in various chemical systems, including gaseous reactions, acid-base equilibria, and complexation reactions. The calculation of equilibrium concentrations often involves a combination of equilibrium constants, such as Ksp, and stoichiometric relationships from the balanced chemical equation. Mastery of this concept is crucial for chemists when predicting reaction extents, optimizing product yields, and designing chemical processes.

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Most popular questions from this chapter

In an attempted determination of the solubility product constant of \(\mathrm{Tl}_{2} \mathrm{~S}\), the solubility of this compound in pure \(\mathrm{CO}_{2}\) free water was determined as \(2.0 \times 10^{-6} \mathrm{M}\). Assume that the dissolved sulphide hydrolyses almost completely to \(\mathrm{HS}^{-}\) and that the further hydrolysis to \(\mathrm{H}_{2} \mathrm{~S}\) can be neglected, what is the computed \(K_{\mathrm{sp}}\) ? For \(\mathrm{H}_{2} \mathrm{~S}, K_{\mathrm{al}}=1.4 \times 10^{-7}, K_{\mathrm{a} 2}=1.0 \times 10^{-14}\) (a) \(6.4 \times 10^{-23}\) (b) \(1.6 \times 10^{-23}\) (c) \(3.2 \times 10^{-17}\) (d) \(3.2 \times 10^{-24}\)

\(\begin{array}{ll}\text { The } \text { equilibrium } & \text { carbonate } \text { ion }\end{array}\) concentration after equal volumes of \(0.7 \mathrm{M}-\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(0.7 \mathrm{M}-\mathrm{HCl}\) solutions are mixed, is \(\left(K_{\mathrm{al}}\right.\) and \(K_{\mathrm{a} 2}\) for \(\mathrm{H}_{2} \mathrm{CO}_{3}\) are \(4.9 \times 10^{-6}\) and \(4.0 \times 10^{-11}\), respectively) (a) \(0.7 \mathrm{M}\) (b) \(0.35 \mathrm{M}\) (c) \(0.002 \mathrm{M}\) (d) \(0.001 \mathrm{M}\)

The correct order of increasing \(\left[\mathrm{OH}^{-}\right]\) in the following aqueous solution is (a) \(0.01 \mathrm{M}-\mathrm{NaHCO}_{3}<0.01 \mathrm{M}-\mathrm{NaCN}\) \(<0.01 \mathrm{M}-\mathrm{KCl}\) (b) \(0.01 \mathrm{M}-\mathrm{KCl}<0.01 \mathrm{M}-\mathrm{NaCN}\) \(<0.01 \mathrm{M}-\mathrm{NaHCO}_{3}\) (c) \(0.01 \mathrm{M}-\mathrm{KCl}<0.01 \mathrm{M}-\mathrm{NaHCO}_{3}\) \(<0.01 \mathrm{M}-\mathrm{NaCN}\) (d) \(0.01 \mathrm{M}-\mathrm{NaCN}<0.01 \mathrm{M}-\mathrm{KCl}\) \(<0.01 \mathrm{M}-\mathrm{NaHCO}_{3}\)

The solubility of \(\mathrm{Li}_{3} \mathrm{Na}_{3}\left(\mathrm{AlF}_{6}\right)_{2}\) is \(0.0744 \mathrm{~g}\) per \(100 \mathrm{ml}\) at \(298 \mathrm{~K}\). Calculate the solubility product of the salt. (Atomic masses: \(\mathrm{Li}=7, \mathrm{Na}=23, \mathrm{Al}=27, \mathrm{~F}=19)\) (a) \(2.56 \times 10^{-22}\) (b) \(2 \times 10^{-3}\) (c) \(7.46 \times 10^{-19}\) (d) \(3.46 \times 10^{-12}\)

An amount of \(0.1\) mole of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) \(\left(K_{\mathrm{b}}=5 \times 10^{-4}\right)\) is mixed with \(0.08\) mole of \(\mathrm{HCl}\) and diluted to one litre. What will be the \(\mathrm{H}^{+}\) concentration in the solution? (a) \(1.25 \times 10^{-4} \mathrm{M}\) (b) \(8 \times 10^{-11} \mathrm{M}\) (c) \(1.6 \times 10^{-11} \mathrm{M}\) (d) \(2 \times 10^{-3} \mathrm{M}\)

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