Question

Consider the reaction \(10_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}(a q) ; K_{c}=3.5 \times 10^{-2} .\) If you start with 25.0 \(\mathrm{mL}\) of a 0.905 \(\mathrm{M}\) solution of \(\mathrm{NaIO}_{4},\) and then dilute it with water to 500.0 \(\mathrm{mL}\) , what is the concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}\) at equilibrium?

Short answer

At equilibrium, the concentration of \(\mathrm{H}_{4}\mathrm{IO}_{6}^{-}\) is approximately \(5.4 \times 10^{-12} \, \mathrm{M}\).

Step-by-step solution

Calculate the initial concentration of \(\mathrm{IO}_{4}^{-}\) after dilution

Use the dilution formula: \(c_1V_1 = c_2V_2\): Here, \(c_1\) = initial concentration of \(\mathrm{NaIO}_{4}\) = 0.905 M \(V_1\) = initial volume of \(\mathrm{NaIO}_{4}\) solution = 25.0 mL \(V_2\) = final volume of solution after dilution = 500.0 mL We need to find \(c_2\), the concentration of \(\mathrm{IO}_{4}^{-}\) after dilution. \[c_2 = \frac{c_1V_1}{V_2}\] Plug in the values to calculate \(c_2\): \[c_2 = \frac{(0.905)(25.0)}{500.0}\] \[c_2 = 0.0452 \, \mathrm{M}\]

Set up an ICE table

Since we need to find out the concentrations of species at equilibrium, we will use an ICE table to keep track of the changes in concentrations throughout the reaction. Assume that the change in \(\mathrm{IO}_{4}^{-}\) concentration is denoted as \(x\). Recall that the stoichiometry of the balanced equation is important for the ICE table. | | \(\mathrm{IO}_{4}^{−}\) (aq) | \(2 \, \mathrm{H}_{2}\mathrm{O}\) (l) | \(\mathrm{H}_{4}\mathrm{IO}_{6}^{-}\) (aq) | |-------|------------------|-----------------|--------------------| |Initial| 0.0452 M | not needed | 0 | | Change| -\(10x\) | -- | \(x\) | |Equilibrium| 0.0452-10x | not needed | \(x\) | Since water doesn't appear in the expression for \(K_c\), its concentration is not needed in the ICE table. \(\DeclareMathOperator{\M}{\:M}\) \(\)

Write the expression for \(K_c\) and plug in ICE table values

In this reaction, the given equilibrium constant is: \[K_c = 3.5 \times 10^{-2}\] Now write the expression for \(K_c\) based on the products and reactants of the reaction: \[K_c = \frac{[\mathrm{H}_{4}\mathrm{IO}_{6}^{-}]}{[\mathrm{IO}_{4}^{-}]^{10}}\] Substitute the equilibrium concentrations from the ICE table into the equation: \[K_c = \frac{x}{(0.0452 - 10x)^{10}}\] Now, plug in the value of \(K_c\) and solve for \(x\): \[3.5 \times 10^{-2} = \frac{x}{(0.0452 - 10x)^{10}}\]

Solve for \(x\) to find the concentration of \(\mathrm{H}_{4}\mathrm{IO}_{6}^{-}\)

In this step, we may assume that \(10x \ll 0.0452\), as the number \(10\) has a significant impact raised to high powers. This means that we can approximate \(0.0452 - 10x ≈ 0.0452\), so the equation then becomes: \[3.5 \times 10^{-2} = \frac{x}{(0.0452 )^{10}}\] \[x = 3.5 \times 10^{-2} \cdot (0.0452 )^{10}\] \[x ≈ 5.4 \times 10^{-12}\] Since \(x\) represents the equilibrium concentration of \(\mathrm{H}_{4}\mathrm{IO}_{6}^{-}\): \[[\mathrm{H}_{4}\mathrm{IO}_{6}^{-}]_{eq} ≈ 5.4 \times 10^{-12} \M\] At equilibrium, the concentration of \(\mathrm{H}_{4}\mathrm{IO}_{6}^{-}\) is approximately \(5.4 \times 10^{-12} \, \mathrm{M}\).

Key concepts

Dilution Calculation

Before exploring the equilibrium concentration, it's vital to understand how dilution calculations are handled. When a solution is diluted, its concentration decreases, but the moles of solute remain constant. This principle can be mathematically represented with the formula:

• \(c_1V_1 = c_2V_2\)

In this equation, \(c_1\) is the initial concentration of the solute, \(V_1\) is the initial volume, \(c_2\) is the final concentration after dilution, and \(V_2\) is the final volume. By rearranging this formula, one can solve for any variable in the equation depending on the given parameters.

For example, suppose we begin with a 25.0 mL solution of 0.905 M \( ext{NaIO}_4\) and dilute it to 500.0 mL. We can determine the new concentration using:

• \(c_2 = \frac{c_1V_1}{V_2}= \frac{(0.905)(25.0)}{500.0}\)

The post-dilution concentration \(c_2\) results in 0.0452 M. This calculated concentration becomes crucial for subsequent steps in multi-part chemical equilibrium calculations.

ICE Table Method

To calculate equilibrium concentrations, chemists often use the ICE table method. ICE stands for Initial, Change, and Equilibrium stages in a chemical reaction. This format organizes the initial concentrations, the changes that occur as the reaction progresses, and the final equilibrium concentrations.

Consider a reaction like \(10_{4}^{-} + 2 \, \mathrm{H}_{2} ext{O} \rightleftharpoons \, \mathrm{H}_{4} ext{IO}_{6}^{-}\). An ICE table is prepared as follows:

• Initially: set known initial concentrations.
• Change: represent changes in concentration with variables \(x\), accounting for stoichiometry.
• Equilibrium: calculate final concentrations using initial values and changes.

For our specific example, the initial concentration of \(\text{IO}_4^+\) is 0.0452 M. Given stoichiometry, express the change of \(x\) as \(-10x\) for \(\text{IO}_4^-\) losses, and \(+x\) for \(\text{H}_4\text{IO}_6^-\) gains. The ICE table thus predicts how each species' concentration shifts toward equilibrium.

Equilibrium Constant (Kc)

The equilibrium constant, \(K_c\), is crucial in predicting the direction and extent of a reaction. It is derived from the concentrations of products and reactants at equilibrium. For any given chemical reaction, it is defined by:

• \[ K_c = \frac{\text{[Products]}}{\text{[Reactants]}} \]

These concentrations are raised to the power of their stoichiometric coefficients from the balanced chemical equation.

In our example reaction, the expression for \(K_c\) is:

• \[K_c = \frac{[\text{H}_4\text{IO}_6^-]}{[\text{IO}_4^-]^{10}}\]

The value of \(K_c = 3.5 \times 10^{-2}\) helps us predict how reactants convert to products and vice versa. Given the high exponent and the small \(K_c\), we expect minimal progression toward products under standard conditions, resulting in tiny equilibrium concentrations.

Chemical Equilibrium

Chemical equilibrium is an essential concept where reactions occur in both directions at equal rates, reaching a state where concentrations remain constant over time. During equilibrium:

• All substances maintain a steady concentration.
• Forward and reverse reaction rates match.
• The system remains dynamic; individual molecules still react.

For the reaction \(10_{4}^{-}\) and \(\text{H}_2\text{O}\) forming \(\text{H}_4\text{IO}_6^-\), equilibrium implies constant concentrations of all species once established. Using our understanding of initial concentrations and changes via ICE tables and calculations involving \(K_c\), we determine these equilibrium concentrations practically, showing how reactants and products stabilize over time.

Le Chatelier's Principle

Le Chatelier's Principle explains how a system at equilibrium responds to changes in concentration, pressure, or temperature. The principle posits that when a change is applied, the system shifts to counteract the disturbance and restore equilibrium.

In the context of our example reaction, if we alter reactant or product concentrations:

• Adding more \(\text{IO}_4^-\) shifts equilibrium towards products to reduce the added reactant's effect.
• Increasing \(\text{H}_4\text{IO}_6^-\) would shift equilibrium back towards \(\text{IO}_4^-\) to counter the disturbance.

Understanding this principle allows chemists to manipulate reactions favorably by intentionally changing conditions, either enhancing or suppressing particular reaction outcomes. It's a balancing act nature performs, leveraging reaction dynamics to achieve new equilibrium states that counteract applied changes.