Calculating PbCO3 Solubility At 25°C In Water And Pb(NO3)2 Solution

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Introduction

In the realm of chemistry, understanding the solubility of compounds is crucial for predicting their behavior in various solutions. Solubility, in simple terms, refers to the maximum amount of a solute that can dissolve in a given amount of solvent at a specific temperature. This property is governed by the solubility product constant (KspK_{sp}), which represents the equilibrium constant for the dissolution of a sparingly soluble salt. In this comprehensive guide, we will delve into the intricate process of calculating the solubility of lead carbonate ($PbCO_3$) at $25^{\circ}C$ in two distinct scenarios: pure water and a $0.0010 M Pb(NO_3)_2$ solution. We will explore the underlying principles, step-by-step calculations, and the significance of the common ion effect. By the end of this guide, you will gain a solid understanding of solubility calculations and their practical applications in chemistry.

Lead carbonate ($PbCO_3$) is a sparingly soluble salt, meaning it dissolves only to a limited extent in water. The dissolution process can be represented by the following equilibrium:

PbCO3(s)⇌Pb2+(aq)+CO32−(aq)PbCO_3(s) \rightleftharpoons Pb^{2+}(aq) + CO_3^{2-}(aq)

The solubility product constant (KspK_{sp}) for this equilibrium is defined as:

Ksp=[Pb2+][CO32−]K_{sp} = [Pb^{2+}][CO_3^{2-}]

where [$Pb^{2+}$] and [$CO_3^{2-}$] represent the molar concentrations of lead(II) ions and carbonate ions, respectively, at equilibrium. The value of $K_{sp}$ is temperature-dependent and can be found in standard reference tables or databases like the ALEKS Data tab, as mentioned in the prompt. This constant is our key to unlocking the solubility of lead carbonate under different conditions.

To accurately calculate the solubility, we must consider not only the $K_{sp}$ value but also the presence of other ions in the solution that might influence the equilibrium. This is where the common ion effect comes into play. The common ion effect describes the decrease in the solubility of a sparingly soluble salt when a soluble salt containing a common ion is added to the solution. In our case, the presence of lead(II) ions from the $Pb(NO_3)2$ solution will affect the solubility of $PbCO_3$. Understanding this effect is crucial for predicting the behavior of lead carbonate in various environments, from natural water systems to industrial processes. So, let's embark on this journey of solubility calculations, where we'll unravel the interplay of equilibrium, the $K{sp}$, and the common ion effect.

Solubility Calculation in Pure Water

Determining the $K_{sp}$ Value

The first step in calculating the solubility of $PbCO_3$ in pure water is to determine the $K_{sp}$ value at $25^{\circ}C$. As specified in the prompt, we will refer to the ALEKS Data tab for this information. Let's assume, for the sake of this example, that the $K_{sp}$ value for $PbCO_3$ at $25^{\circ}C$ is found to be $1.5 \times 10^{-13}$. This value is a cornerstone for our calculations, representing the maximum product of the ion concentrations in a saturated solution.

Setting up the Equilibrium Expression

As previously stated, the dissolution equilibrium for $PbCO_3$ is:

PbCO3(s)⇌Pb2+(aq)+CO32−(aq)PbCO_3(s) \rightleftharpoons Pb^{2+}(aq) + CO_3^{2-}(aq)

To quantify the solubility, let's define 's' as the molar solubility of $PbCO_3$ in pure water. This means that for every mole of $PbCO_3$ that dissolves, one mole of $Pb^{2+}$ ions and one mole of $CO_3^{2-}$ ions are produced in the solution. Therefore, at equilibrium:

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

[$CO_3^{2-}$] = s

Applying the $K_{sp}$ Expression

Now, we can substitute these concentrations into the $K_{sp}$ expression:

Ksp=[Pb2+][CO32−]=(s)(s)=s2K_{sp} = [Pb^{2+}][CO_3^{2-}] = (s)(s) = s^2

Solving for Solubility (s)

We now have a simple equation relating the $K_{sp}$ to the solubility 's':

s2=Ksps^2 = K_{sp}

To find 's', we take the square root of both sides:

s=Ksps = \sqrt{K_{sp}}

Substituting the assumed $K_{sp}$ value of $1.5 \times 10^{-13}$:

s=1.5×10−13≈1.2×10−6Ms = \sqrt{1.5 \times 10^{-13}} \approx 1.2 \times 10^{-6} M

Rounding to Significant Figures

The prompt instructs us to round the answer to two significant figures. Therefore, the solubility of $PbCO_3$ in pure water at $25^{\circ}C$ is approximately $1.2 \times 10^{-6} M$. This result indicates the extremely low solubility of lead carbonate in pure water, highlighting its classification as a sparingly soluble salt. This calculated value serves as a baseline for comparison when we explore the solubility in the presence of a common ion. Understanding the solubility in pure water is essential as it provides a reference point to assess how other factors, such as the common ion effect, can alter the dissolution equilibrium.

Solubility Calculation in a $0.0010 M Pb(NO_3)_2$ Solution

Understanding the Common Ion Effect

The presence of $Pb(NO_3)_2$ in the solution introduces the common ion effect, which, as discussed earlier, will decrease the solubility of $PbCO_3$. $Pb(NO_3)_2$ is a soluble salt that dissociates completely in water, providing $Pb^{2+}$ ions to the solution. This increase in the concentration of $Pb^{2+}$ ions will shift the dissolution equilibrium of $PbCO_3$ to the left, according to Le Chatelier's principle, thus reducing the amount of $PbCO_3$ that can dissolve.

Initial Concentrations

Before considering the dissolution of $PbCO_3$, let's determine the initial concentrations of ions in the $0.0010 M Pb(NO_3)_2$ solution. Since $Pb(NO_3)_2$ is a strong electrolyte, it dissociates completely as follows:

Pb(NO3)2(s)→Pb2+(aq)+2NO3−(aq)Pb(NO_3)_2(s) \rightarrow Pb^{2+}(aq) + 2NO_3^{-}(aq)

Therefore, a $0.0010 M Pb(NO_3)_2$ solution will produce $0.0010 M$ of $Pb^{2+}$ ions and $0.0020 M$ of $NO_3^{-}$ ions. The initial concentration of $Pb^{2+}$ is crucial for understanding how the common ion effect will influence the solubility of $PbCO_3$.

Setting up the Equilibrium Expression with the Common Ion

Now, let's consider the dissolution of $PbCO_3$ in this solution. Again, we define 's' as the molar solubility of $PbCO_3$ in the $0.0010 M Pb(NO_3)_2$ solution. The equilibrium remains the same:

PbCO3(s)⇌Pb2+(aq)+CO32−(aq)PbCO_3(s) \rightleftharpoons Pb^{2+}(aq) + CO_3^{2-}(aq)

However, the initial concentration of $Pb^{2+}$ is no longer zero. We have $0.0010 M$ of $Pb^{2+}$ from the $Pb(NO_3)_2$ solution. At equilibrium, the concentrations will be:

[$Pb^{2+}$] = 0.0010 + s

[$CO_3^{2-}$] = s

Applying the $K_{sp}$ Expression

Substituting these equilibrium concentrations into the $K_{sp}$ expression:

Ksp=[Pb2+][CO32−]=(0.0010+s)(s)K_{sp} = [Pb^{2+}][CO_3^{2-}] = (0.0010 + s)(s)

Solving for Solubility (s) – Approximation Method

We now have a quadratic equation to solve for 's':

1.5×10−13=(0.0010+s)(s)1. 5 \times 10^{-13} = (0.0010 + s)(s)

However, since $PbCO_3$ is sparingly soluble and the $K_{sp}$ value is very small, we can make an approximation to simplify the calculation. We can assume that 's' is much smaller than 0.0010, so 0.0010 + s ≈ 0.0010. This approximation is valid if the value of 's' we calculate is less than 5% of 0.0010. With this approximation, the equation becomes:

1.5×10−13=(0.0010)(s)1. 5 \times 10^{-13} = (0.0010)(s)

Solving for 's':

s=1.5×10−130.0010=1.5×10−10Ms = \frac{1.5 \times 10^{-13}}{0.0010} = 1.5 \times 10^{-10} M

Verifying the Approximation

Before accepting this result, we must verify our approximation. Is $1.5 \times 10^{-10}$ less than 5% of 0.0010? 5% of 0.0010 is $5 \times 10^{-5}$, and $1.5 \times 10^{-10}$ is indeed much smaller than $5 \times 10^{-5}$, so our approximation is valid. If the approximation wasn't valid, we would need to solve the quadratic equation using the quadratic formula or an iterative method.

Rounding to Significant Figures

Rounding our answer to two significant figures, the solubility of $PbCO_3$ in the $0.0010 M Pb(NO_3)_2$ solution at $25^{\circ}C$ is approximately $1.5 \times 10^{-10} M$.

Comparing Solubilities

Comparing this result to the solubility in pure water ($1.2 \times 10^{-6} M$), we observe a significant decrease in solubility due to the common ion effect. The solubility in the $Pb(NO_3)_2$ solution is several orders of magnitude lower than in pure water, clearly demonstrating the impact of the common ion on the dissolution equilibrium. This difference underscores the importance of considering the solution composition when calculating the solubility of sparingly soluble salts. The common ion effect is a fundamental concept in solubility chemistry, with implications for various applications, including analytical chemistry, environmental science, and pharmaceutical formulations.

Conclusion

In this comprehensive guide, we have successfully calculated the solubility of lead carbonate ($PbCO_3$) at $25^{\circ}C$ in both pure water and a $0.0010 M Pb(NO_3)2$ solution. We began by understanding the fundamental principles of solubility and the significance of the solubility product constant ($K{sp}$). We then meticulously calculated the solubility in pure water, using the $K_{sp}$ value to determine the equilibrium concentrations of $Pb^{2+}$ and $CO_3^{2-}$ ions. This calculation provided a baseline for understanding the inherent solubility of $PbCO_3$.

Furthermore, we delved into the intricacies of the common ion effect, which significantly impacts the solubility of sparingly soluble salts. By introducing $Pb(NO_3)_2$ to the solution, we increased the concentration of the common ion, $Pb^{2+}$, thereby shifting the dissolution equilibrium of $PbCO_3$ to the left and reducing its solubility. We applied an approximation method to simplify the calculation, which proved to be valid in this case, and obtained a solubility value that was several orders of magnitude lower than in pure water.

The results clearly demonstrate the profound influence of the common ion effect on the solubility of $PbCO_3$. This phenomenon has significant implications in various fields. In environmental chemistry, understanding the solubility of lead compounds is crucial for assessing the potential for lead contamination in water systems. In analytical chemistry, the common ion effect can be exploited to control the precipitation of certain ions for separation and analysis. In the pharmaceutical industry, solubility is a critical factor in drug formulation and delivery.

In summary, this guide has provided a step-by-step approach to calculating solubility, highlighting the importance of the $K_{sp}$ value and the common ion effect. By mastering these concepts, you can confidently tackle a wide range of solubility problems and apply this knowledge to real-world scenarios. Solubility calculations are not just theoretical exercises; they are essential tools for understanding and predicting the behavior of chemical compounds in various environments. Understanding these concepts allows chemists and other scientists to manipulate chemical systems for desired outcomes, whether it's designing a new drug, cleaning up environmental contamination, or developing new materials.