/*! 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 102 Calculate the ratio of \(\left[\... [FREE SOLUTION] | 91影视

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Calculate the ratio of \(\left[\mathrm{Ca}^{2+}\right]\) to \(\left[\mathrm{Fe}^{2+}\right]\) in a lake in which the water is in equilibrium with deposits of both \(\mathrm{CaCO}_{3}\) and \(\mathrm{FeCO}_{3}\). Assume that the water is slightly basic and that the hydrolysis of the carbonate ion can therefore be ignored.

Short Answer

Expert verified
The ratio of the concentrations of Ca虏鈦 and Fe虏鈦 ions in the lake is approximately 101.5:1.

Step by step solution

01

Write the chemical equilibrium equations

The chemical equilibrium equations for the dissolution of CaCO鈧 and FeCO鈧 in water are as follows: \(CaCO_{3}(s) \rightleftharpoons Ca^{2+}(aq) + CO_{3}^{2-}(aq)\) \(FeCO_{3}(s) \rightleftharpoons Fe^{2+}(aq) + CO_{3}^{2-}(aq)\)
02

Write the solubility product expressions

The solubility product expressions (Ksp) for the above equilibrium equations would be: \(K_{sp(CaCO_{3})} = [Ca^{2+}][CO_{3}^{2-}]\) \(K_{sp(FeCO_{3})} = [Fe^{2+}][CO_{3}^{2-}]\)
03

Obtain the solubility product constants

The solubility product constants (Ksp) for CaCO鈧 and FeCO鈧 can be found in a reference book or table. For this problem, let's assume the following Ksp values: \(K_{sp(CaCO_{3})} = 3.36 \times 10^{-9}\) \(K_{sp(FeCO_{3})} = 3.31 \times 10^{-11}\)
04

Express both concentrations in terms of a common variable

Let's express the concentrations of both Ca虏鈦 and Fe虏鈦 in terms of the concentration of the carbonate ion (CO鈧兟测伝). Using the solubility product expressions from Step 2, we can write: \([Ca^{2+}] = \frac{K_{sp(CaCO_{3})}}{[CO_{3}^{2-}]}\) \([Fe^{2+}] = \frac{K_{sp(FeCO_{3})}}{[CO_{3}^{2-}]}\)
05

Calculate the ratio of the concentrations

Now we will find the ratio of the concentrations of Ca虏鈦 and Fe虏鈦 ions: \(\frac{[Ca^{2+}]}{[Fe^{2+}]} = \frac{\frac{K_{sp(CaCO_{3})}}{[CO_{3}^{2-}]}}{\frac{K_{sp(FeCO_{3})}}{[CO_{3}^{2-}]}} = \frac{K_{sp(CaCO_{3})}}{K_{sp(FeCO_{3})}}\) Using the Ksp values obtained in Step 3, we can calculate the ratio: \(\frac{[Ca^{2+}]}{[Fe^{2+}]} = \frac{3.36 \times 10^{-9}}{3.31 \times 10^{-11}} = 101.5\) Therefore, the ratio of the concentrations of Ca虏鈦 and Fe虏鈦 ions in the lake is approximately 101.5:1.

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Key Concepts

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

Equilibrium Equations
In chemistry, equilibrium equations are fundamental when analyzing solubility in a solution. These equations represent the state where the forward and reverse processes, such as dissolution and precipitation, are balanced. For instance, the equilibrium equations for calcium carbonate (\(CaCO_3\)) and iron carbonate (\(FeCO_3\)) sharing the same aqueous environment reveal how each compound dissociates into its constituent ions.
The relevant equations are:

  • \(CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)\)
  • \(FeCO_3(s) \rightleftharpoons Fe^{2+}(aq) + CO_3^{2-}(aq)\)
These expressions highlight the ions' distribution in the liquid phase and are essential for predicting the behavior of solutions in various chemical reactions. Each side of the equilibrium depicts a balanced process, demonstrating the continuous interchange between solid and dissolved ions.
Understanding these reactions' equilibrium allows us to predict how changes in conditions, like concentration or temperature, may affect the solubility of each compound in the mixture. This balance is crucial for many natural processes, such as the formation of sedimentary rock layers or the buffering capacity of natural bodies of water.
Solubility Product Constant
The solubility product constant, often denoted as \(K_{sp}\), is an essential parameter in understanding how concentrated a solution can become before precipitating a solid. It specifically applies to sparingly soluble ionic compounds and is derived from the equilibrium expression of the solute.
For each equilibrium equation, the \(K_{sp}\) relates to the concentrations of the dissolved ions. For instance, looking at calcium and iron carbonates, the expressions are:

  • \(K_{sp(CaCO_3)} = [Ca^{2+}][CO_3^{2-}]\)
  • \(K_{sp(FeCO_3)} = [Fe^{2+}][CO_3^{2-}]\)
The values of \(K_{sp}\) can provide insight into how much of each salt can dissolve at equilibrium. These constants are determined experimentally and can vary with temperature. In our example, higher \(K_{sp}\) values signify a greater solubility.
This concept helps chemists understand and calculate concentrations seen in various natural and synthetic environments. By comparing solubility product constants, we can predict which compounds are more likely to precipitate under given conditions, leading us to analyze interactions and transformations effectively.
Concentration Ratio
The concentration ratio of ions in a solution is a critical factor in determining relative solubility and potential precipitation reactions. It is especially useful for deducing relative abundance or scarcity when two or more ionic solids share a common ion.
For the given problem, the concentration ratio of \([Ca^{2+}]\) to \([Fe^{2+}]\) is calculated using their respective \(K_{sp}\) values against the carbonate ion, \([CO_3^{2-}]\). By expressing each concentration in terms of the carbonate ion and simplifying, we find:

  • \(\frac{[Ca^{2+}]}{[Fe^{2+}]} = \frac{K_{sp(CaCO_3)}}{K_{sp(FeCO_3)}}\)
Given that the solubility product constants were:
  • \(K_{sp(CaCO_3)} = 3.36 \times 10^{-9}\)
  • \(K_{sp(FeCO_3)} = 3.31 \times 10^{-11}\)
The calculated ratio indicates \(101.5:1\), meaning calcium is significantly more soluble under these conditions.
This concentration ratio helps in predicting scaling tendencies, potential mineral deposits, and biologically critical processes in natural waters. By understanding how to calculate and interpret this ratio, one gains insights into underlying chemical equilibria and their implications in various fields like geochemistry and environmental sciences.

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Most popular questions from this chapter

Suppose you want to do a physiological experiment that calls for a pH 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid \(\mathrm{H}_{2} \mathrm{X}\) \(\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and a \(1.0 \mathrm{M}\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH} 6.50 ?\) (Ignore any volume change.)

Explain why a mixture of \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{COONa}\) can act as a buffer while a mixture of \(\mathrm{HCl}\) and \(\mathrm{NaCl}\) cannot.

Calculate the solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in \(0.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\).

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