Problem 121 - Chemistry The Science in Context

Question

A puddle of coastal seawater, caught in a depression formed by some coastal rocks at high tide, begins to evaporate on a hot summer day as the tide goes out. If the volume of the puddle decreases to \(23 \%\) of its initial volume, what is the concentration of \(\mathrm{Na}^{+}\) after evaporation if initially it was \(0.449 M ?\)

Step-by-Step Solution

Answer

Answer: The final concentration of sodium ions after the evaporation is approximately 1.95 M.

Step 1: Write down the relation between initial and final concentrations/volumes

We know that the product of the initial concentration and volume is equal to the final concentration and volume. So, we have: initial_concentration × initial_volume = final_concentration × final_volume

Step 2: Calculate the final volume

The final volume is 23% of the initial volume. We can write this as: final_volume = 0.23 × initial_volume

Step 3: Substitute the initial concentration and final volume in the relation

Given that the initial concentration of Na+ is 0.449 M, we can now substitute the values into our equation. 0.449 × initial_volume = final_concentration × (0.23 × initial_volume)

Step 4: Solve the equation for the final concentration

In order to solve for the final concentration, we can simply divide both sides by the product of 0.23 and the initial volume. final_concentration = \frac{0.449 \times initial_volume}{0.23 \times initial_volume} Since we only need the final concentration, the initial_volume terms will cancel out: final_concentration = \frac{0.449}{0.23}

Step 5: Calculate the final concentration

Finally, we can calculate the final concentration of Na+ ions: final_concentration = \frac{0.449}{0.23} = 1.95217391 M

Step 6: Write the answer

The concentration of Na+ ions after evaporation of the puddle is approximately 1.95 M.

Key Concepts

Evaporation

Evaporation is the process where liquid turns into vapor. It typically occurs at the surface of the liquid and does not require the liquid to reach its boiling point. During a hot day, evaporation can happen quite rapidly.
In situations like a puddle of seawater on a coastal rock, evaporation leads to a reduction in water volume, leaving behind the dissolved salts. This makes any remaining solution more concentrated, as the same amount of solute (dissolved substance) is surrounded by less solvent (liquid).
Understanding evaporation is crucial in many fields, such as meteorology, oceanography, and environmental science, where it's important to predict how substances might concentrate or dilute due to water loss.

Sodium Ion Concentration

Sodium ions, denoted as Na⁺, are a crucial component of seawater's salinity. These ions attract water molecules, which is why they're often involved in processes affecting water balance.

• In a solution, concentration refers to the amount of a particular solute in a given volume of solvent.
• As a solution evaporates and volume decreases, the concentration of sodium ions increases if no ions are lost.

This principle is often observed in coastal regions, where evaporation rates are high. Sodium ion concentration affects various environmental and biological processes, influencing factors such as ocean currents and marine life.

Concentration Calculations

To find the concentration after evaporation, we must understand the relationship between initial and final states of a solution. Concentration is often expressed in molarity (M), which stands for moles per liter.
The key calculation follows the principle of conservation of moles. If you start with an initial concentration ( \( C_i \)) and an initial volume ( \( V_i \)), and end with a final concentration ( \( C_f \)) and final volume ( \( V_f \)), the equation is:

• \( C_i \times V_i = C_f \times V_f \)

To find the final concentration of Na⁺ after evaporation:

• Given the initial concentration ( \( C_i = 0.449 \) M)
• Remember the final volume is 23% of the initial volume:
• \( V_f = 0.23 \times V_i \)
• The formula simplifies as the initial volumes cancel out: \( C_f = \frac{0.449}{0.23} \)

This gives a final sodium ion concentration approximated as 1.95 M. Accurate calculations can predict how factors like concentration will change based on varying conditions, making it a valuable tool in chemistry.