Calculating Solubility Of Nickel(II) Hydroxide At 25°C In Pure Water And NaOH Solution

Understanding the solubility of chemical compounds is crucial in various scientific disciplines, ranging from environmental chemistry to pharmaceutical science. Solubility, defined as the maximum amount of a solute that can dissolve in a given solvent at a specific temperature, is governed by the solubility product constant, often denoted as Ksp. This article aims to provide a detailed explanation of how to calculate the solubility of Nickel(II) hydroxide ($Ni(OH)_2$) in two distinct scenarios: pure water and a 0.0070 M NaOH solution, at a temperature of 25°C. We will utilize the $K_{sp}$ data, commonly found in chemistry data tables, to perform these calculations and round our answers to two significant digits. This comprehensive guide is designed to assist students, researchers, and anyone interested in grasping the principles behind solubility calculations and the common ion effect.

Solubility is a fundamental concept in chemistry, reflecting the extent to which a substance dissolves in a solvent. For sparingly soluble compounds like $Ni(OH)_2$, the dissolution process is an equilibrium reaction. When a solid ionic compound is introduced into water, it dissociates into its constituent ions. For instance, $Ni(OH)_2$ dissociates into $Ni^{2+}$ and $OH^-$ ions. The solubility product constant ($K_{sp}$) is the equilibrium constant for this dissolution reaction. It represents the product of the ion concentrations at saturation, each raised to the power of its stoichiometric coefficient in the balanced dissolution equation. The $K_{sp}$ value is temperature-dependent and serves as a measure of a compound's solubility; a lower $K_{sp}$ indicates lower solubility, and vice versa. Understanding the $K_{sp}$ is essential for predicting whether a precipitate will form when solutions are mixed or for calculating the solubility of a compound under different conditions. In the following sections, we will delve into how to use the $K_{sp}$ of $Ni(OH)_2$ to calculate its solubility in both pure water and a solution containing a common ion.

To determine the solubility of $Ni(OH)_2$ in pure water, we first need to establish the equilibrium expression for its dissolution. The dissolution of $Ni(OH)_2$ in water can be represented by the following equation:

$Ni(OH)_2(s) ightleftharpoons Ni^{2+}(aq) + 2OH^-(aq)$

The solubility product expression ($K_{sp}$) for this reaction is:

$K_{sp} = [Ni^{2+}][OH^-]^2$

Let 's' represent the molar solubility of $Ni(OH)_2$ in water. This means that for every mole of $Ni(OH)_2$ that dissolves, 's' moles of $Ni^{2+}$ and '2s' moles of $OH^-$ are produced. Thus, at equilibrium:

$[Ni^{2+}] = s$

$[OH^-] = 2s$

Substituting these concentrations into the $K_{sp}$ expression, we get:

$K_{sp} = (s)(2s)^2 = 4s^3$

From the ALEKS Data tab or standard chemistry resources, the $K_{sp}$ of $Ni(OH)_2$ at 25°C is approximately $2.0 × 10^{-15}$. Plugging this value into the equation, we have:

$2.0 × 10^{-15} = 4s^3$

Solving for 's':

$s^3 = (2.0 × 10^{-15}) / 4 = 5.0 × 10^{-16}$

$s =

∛(5.0 × 10^{-16}) ≈ 8.0 × 10^{-6} M$

Therefore, the solubility of $Ni(OH)_2$ in pure water at 25°C is approximately $8.0 × 10^{-6}$ M, rounded to two significant digits. This calculation demonstrates the direct relationship between the $K_{sp}$ and the solubility of a sparingly soluble salt in pure water. The molar solubility is quite low, indicating that $Ni(OH)_2$ is indeed sparingly soluble. In the next section, we will explore how the presence of a common ion affects the solubility of $Ni(OH)_2$.

Now, let's calculate the solubility of $Ni(OH)_2$ in a 0.0070 M NaOH solution. The presence of NaOH introduces a common ion, $OH^-$, which affects the equilibrium of the $Ni(OH)_2$ dissolution. This is known as the common ion effect, which generally decreases the solubility of a sparingly soluble salt. The dissolution equilibrium remains:

$Ni(OH)_2(s) ightleftharpoons Ni^{2+}(aq) + 2OH^-(aq)$

However, we now have an initial concentration of $OH^-$ from the NaOH solution. Let 's' be the molar solubility of $Ni(OH)_2$ in the NaOH solution. At equilibrium:

$[Ni^{2+}] = s$

$[OH^-] = 2s + 0.0070 M$

The $K_{sp}$ expression is still:

$K_{sp} = [Ni^{2+}][OH^-]^2$

Substituting the equilibrium concentrations:

$2.0 × 10^{-15} = (s)(2s + 0.0070)^2$

Since the $K_{sp}$ is very small, we can assume that 's' is much smaller than 0.0070, and thus, 2s can be neglected in the sum (2s + 0.0070). This simplifies the equation to:

$2.0 × 10^{-15} = s(0.0070)^2$

Solving for 's':

$s = (2.0 × 10^{-15}) / (0.0070)^2$

$s ≈ 4.1 × 10^{-11} M$

Therefore, the solubility of $Ni(OH)_2$ in a 0.0070 M NaOH solution at 25°C is approximately $4.1 × 10^{-11}$ M, rounded to two significant digits. This result is significantly lower than the solubility in pure water, illustrating the common ion effect. The presence of the common ion ($OH^-$) from NaOH has drastically reduced the solubility of $Ni(OH)_2$. This principle is widely applied in various chemical processes, such as controlling the precipitation of metal hydroxides in wastewater treatment.

Comparing the solubility of $Ni(OH)_2$ in pure water ($8.0 × 10^{-6}$ M) and in the 0.0070 M NaOH solution ($4.1 × 10^{-11}$ M) clearly demonstrates the impact of the common ion effect. The solubility in the NaOH solution is several orders of magnitude lower than in pure water. This reduction in solubility occurs because the presence of $OH^-$ ions from NaOH shifts the dissolution equilibrium of $Ni(OH)_2$ to the left, according to Le Chatelier's principle. Le Chatelier's principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. In this case, the stress is the addition of $OH^-$ ions, and the system responds by reducing the dissolution of $Ni(OH)_2$, thereby decreasing the concentration of $Ni^{2+}$ ions in the solution. This phenomenon is not only a fundamental concept in chemistry but also has practical applications in various fields, such as controlling the precipitation of metal ions in industrial processes and analytical chemistry. Understanding the common ion effect allows chemists to manipulate solubility and control the formation of precipitates, which is crucial in many chemical reactions and separations.

The principles governing the solubility of $Ni(OH)_2$ and the common ion effect have significant practical implications and applications across various fields. In environmental chemistry, understanding the solubility of metal hydroxides is essential for predicting the fate and transport of heavy metals in aquatic systems. For instance, the solubility of $Ni(OH)_2$ can influence the concentration of nickel ions in natural waters, which can have toxic effects on aquatic life. Controlling the pH and the concentration of common ions can be a strategy for mitigating metal contamination in water bodies. In industrial wastewater treatment, the common ion effect is often utilized to precipitate metal ions as hydroxides, thereby removing them from the wastewater before discharge. By adding a base like NaOH, the concentration of $OH^-$ ions is increased, reducing the solubility of metal hydroxides and promoting their precipitation. This process is a cost-effective and efficient method for treating industrial effluents. In analytical chemistry, the solubility product and the common ion effect are crucial for gravimetric analysis, where the quantitative determination of an analyte is based on the precipitation of a compound with known stoichiometry. By carefully controlling the conditions, such as pH and the concentration of common ions, the precipitation can be made selective and quantitative. Furthermore, in the pharmaceutical industry, the solubility of drug compounds is a critical factor in drug formulation and bioavailability. Understanding and manipulating solubility using principles like the common ion effect can improve drug absorption and efficacy. Thus, the knowledge of solubility and its influencing factors is vital in various scientific and industrial applications.

In conclusion, we have thoroughly explored the calculation of the solubility of $Ni(OH)_2$ at 25°C in both pure water and a 0.0070 M NaOH solution. By utilizing the solubility product constant ($K_{sp}$), we determined that the solubility in pure water is approximately $8.0 × 10^{-6}$ M, while the solubility in the NaOH solution is significantly lower, at approximately $4.1 × 10^{-11}$ M. This stark difference exemplifies the common ion effect, where the presence of a common ion ($OH^-$ in this case) reduces the solubility of a sparingly soluble salt. These calculations highlight the importance of understanding equilibrium principles and their application in predicting and controlling solubility. The practical implications of these concepts are vast, spanning environmental science, industrial wastewater treatment, analytical chemistry, and the pharmaceutical industry. Mastery of solubility calculations and the common ion effect is essential for anyone working in these fields, as it provides a powerful tool for manipulating chemical processes and solving real-world problems. This guide serves as a comprehensive resource for students, researchers, and professionals seeking to deepen their understanding of solubility and its applications.