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A \(50.0-\mathrm{mL}\) sample of \(0.00200 \mathrm{M} \mathrm{AgNO}_{3}\) is added to \(50.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{NaIO}_{3} .\) What is the equilibrium concentration of \(\mathrm{Ag}^{+}\) in solution? \(\left(K_{\mathrm{sp}}\right.\) for \(\mathrm{AgIO}_{3}\) is \(\left.3.0 \times 10^{-8} .\right)\)

Short Answer

Expert verified
The equilibrium concentration of Ag鈦 in the solution is \(7.50 \times 10^{-8} \mathrm{mol/L}\).

Step by step solution

01

Write the balanced chemical equation

Write the balanced equation for the reaction between AgNO鈧 and NaIO鈧 which results in the formation of AgIO鈧. AgNO鈧(aq) + NaIO鈧(aq) 鈫 AgIO鈧(s) + NaNO鈧(aq)
02

Set up an ICE table

The next step is to set up an ICE (Initial, Change, Equilibrium) table to relate the initial and equilibrium concentrations of the species. The change in concentration of each species will depend on the stoichiometry of the balanced equation. | | AgNO鈧 | NaIO鈧 | AgIO鈧 | NaNO鈧 | |-----|-------|-------|-------|-------| | I | 0.1 mmol | 0.5 mmol | 0 | 0 | | C | -0.1 mmol | -0.1 mmol | +0.1 mmol | +0.1 mmol | | E | 0 | 0.4 mmol | 0.1 mmol | 0.1 mmol | Here, MM stands for millimoles. Since we are given that 50 mL of 0.00200 M AgNO鈧 is mixed with 50 mL of 0.0100 M NaIO鈧, initial concentrations of those can be calculated from the molarity multiplied by the volumes. Initial millimoles of AgNO鈧 = 0.00200 mol/L * 50 mL * (1L/1000 mL) = 0.1 mmol Initial millimoles of NaIO鈧 = 0.0100 mol/L * 50 mL * (1L/1000 mL) = 0.5 mmol
03

Write Ksp expression for AgIO鈧

Write the solubility product expression, Ksp, for AgIO鈧 based on the balanced equation in Step 1. Ksp = [Ag鈦篯[IO鈧冣伝] Since the Ksp value is very small (3.0 x 10鈦烩伕), we can make an assumption that the change in the concentration of IO鈧冣伝 can be neglected. This is because only a tiny amount of AgIO鈧 will dissolve in the solution at equilibrium, and thus the change in IO鈧冣伝 concentration will be very small. This assumption simplifies the Ksp expression and allows us to solve for the equilibrium concentration of Ag鈦 more easily. Our new Ksp expression will be: Ksp = [Ag鈦篯(0.4 mmol / 0.1 L)
04

Solve for equilibrium concentration of Ag鈦

Use the Ksp value given (3.0 x 10鈦烩伕) to solve for the equilibrium concentration of Ag鈦. 3.0 x 10鈦烩伕 = [Ag鈦篯(0.4 mmol / 0.1 L) We can now solve for [Ag鈦篯: [Ag鈦篯 = (3.0 x 10鈦烩伕) / (0.4 mmol / 0.1 L) = 7.50 x 10鈦烩伕 mol/L Hence, the equilibrium concentration of Ag鈦 in the solution is \(7.50 \times 10^{-8} \mathrm{mol/L}\).

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Key Concepts

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

Solubility Product Constant (Ksp)
The Solubility Product Constant, commonly known as Ksp, is a fundamental concept in chemical equilibrium, especially in sparingly soluble salts. Ksp represents the level at which a solute will dissolve in a solution to form a saturated solution. It is the product of the concentrations of the ions of the salt, each raised to the power of their stoichiometric coefficients.
For the compound AgIO鈧, the Ksp is given by the expression:
  • \( K_{sp} = [\text{Ag}^+][\text{IO}_3^-] \)
In our example, calculating the Ksp involves finding the equilibrium concentrations of these ions in the solution.
The Ksp value allows us to determine the concentration of ions that can exist in solution before precipitation occurs. A low Ksp value, like \(3.0 \times 10^{-8}\) for AgIO鈧, indicates that only a tiny concentration of AgIO鈧 can dissolve in water before it starts precipitating. Understanding and applying Ksp is vital for solving problems related to precipitation and solubility.
ICE Table
An ICE table is a useful tool for tracking the changes in concentrations or pressures for chemical species in equilibrium processes. ICE stands for Initial, Change, and Equilibrium.
To construct an ICE table, begin with the initial concentrations of the reactants and products. Then, determine how these concentrations change during the reaction, and finally, list the equilibrium concentrations.
In our scenario:
  • Initial millimoles of AgNO鈧 and NaIO鈧 are calculated using the formula: \( \text{Molarity} \times \text{Volume} \).
  • The change in concentration is determined by the stoichiometry of the balanced equation.
  • Finally, we calculate the equilibrium concentrations by adding the change to the initial values. For AgNO鈧, the concentration drops to zero, while for NaIO鈧, it decreases slightly, reflecting the ongoing reaction.
ICE tables simplify the calculation process and allow us to visualize the progression of the reaction to equilibrium, thus making it easier to relate changes in concentration back to stoichiometry and Ksp.
Stoichiometry
Stoichiometry is the study of the relative quantities of reactants and products in chemical reactions. Understanding stoichiometry allows us to derive the relationships in the balanced chemical equation, such as the one between AgNO鈧 and NaIO鈧 forming AgIO鈧 and NaNO鈧.
Chemical equations must be balanced to reflect the conservation of mass, meaning the number of atoms for each element is equal on both sides of the equation. This principle is crucial when setting up an ICE table, as it dictates the changes in concentrations for each species.
In the given exercise, stoichiometry tells us that each mole of AgNO鈧 reacts with one mole of NaIO鈧 to produce one mole of AgIO鈧 and NaNO鈧.
Tracking these relationships through stoichiometry allows us to accurately calculate the concentrations at each step of the ICE table, crucial for later solving the Ksp equation.
Precipitation Reaction
A precipitation reaction occurs when ions in aqueous solution combine to form an insoluble solid or precipitate. For AgIO鈧, the formation of a solid indicates that the solution has reached saturation and any additional solute will precipitate out of solution.
The supplied Ksp value helps predict at what point precipitation will begin. For AgNO鈧 and NaIO鈧, the mixing initiates a precipitation reaction forming solid AgIO鈧. The resulting change in ion concentrations takes the system to equilibrium. If the calculated ion product is greater than the published Ksp, precipitation will occur until equilibrium is restored.
Understanding precipitation reactions is essential in predicting whether a solution will remain clear or form a solid. It also underlines the importance of the Ksp value in determining the solubility limits of different ion pairs, further aiding in solving complex equilibrium problems in chemistry.

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