Problem 72 - Chemistry: The Central Science

Question

Indicate the concentration of each ion present in the solution formed by mixing (a) \(42.0 \mathrm{~mL}\) of \(0.170 \mathrm{M} \mathrm{NaOH}\) with \(37.6 \mathrm{~mL}\) of \(0.400 \mathrm{M} \mathrm{NaOH}\), (b) \(44.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) with \(25.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{KCl}\), (c) \(3.60 \mathrm{~g} \mathrm{KCl}\) in \(75.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{CaCl}_{2}\) solution. Assume that the volumes are additive.

Step-by-Step Solution

Answer

(a) The concentration of each ion (Na+ and OH-) in the solution after mixing is \(0.278~\mathrm{M}\). (b) Concentrations after mixing: Na+ = \(0.127~\mathrm{M}\), SO42- = \(0.0636~\mathrm{M}\), K+ = \(0.0469~\mathrm{M}\), and Cl- = \(0.0469~\mathrm{M}\). (c) Concentrations after mixing: K+ = \(0.538~\mathrm{M}\), Cl- = \(1.08~\mathrm{M}\), and Ca2+ = \(0.214~\mathrm{M}\).

Step 1: Determine the moles of each ion in the initial solutions

Since both solutions are NaOH, they each have Na+ and OH- ions. To find the moles of ions in the initial solutions, we will multiply the volume of each solution by its corresponding molarity. Moles of Na+ and OH- in \(42.0 \mathrm{~mL}\) of \(0.170 \mathrm{M} \mathrm{NaOH}\) = \(42.0 \times 0.170~\mathrm{mmol}\) Moles of Na+ and OH- in \(37.6 \mathrm{~mL}\) of \(0.400 \mathrm{M} \mathrm{NaOH}\) = \(37.6 \times 0.400~\mathrm{mmol}\)

Step 2: Add up the moles of each ion after mixing

To find the total moles of Na+ and OH- after mixing, we add up the moles calculated in step 1. Total moles of Na+ and OH- after mixing = \((42.0 \times 0.170) + (37.6 \times 0.400)~\mathrm{mmol}\)

Step 3: Calculate the total volume after mixing

We assume that the volumes are additive, so we will add the initial volumes of the two solutions to find the total volume after mixing. Total volume = \(V_{1} + V_{2} = 42.0~\mathrm{mL} + 37.6~\mathrm{mL}\)

Step 4: Find the concentration of each ion

To find the concentration of each ion, we will divide the total moles of each ion (step 2) by the total volume after mixing (step 3). Concentration of Na+ and OH- = \(\frac{(42.0 \times 0.170) + (37.6 \times 0.400)}{42.0 + 37.6}~\mathrm{M}\) Calculating the values above, we obtain: Concentration of Na+ and OH- = \(\frac{(7.14 + 15.04)}{79.6}~\mathrm{M} = 0.278~\mathrm{M}\) Therefore, the concentration of each ion (Na+ and OH-) in the solution after mixing is \(0.278~\mathrm{M}\). (b) Mixing \(44.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) with \(25.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{KCl}\)

Step 1: Determine the moles of each ion in the initial solutions

The solutions have Na+, SO42-, K+ and Cl- ions. To find the moles of ions in the initial solutions, we will multiply the volume of each solution by its corresponding molarity. Moles of Na+ in \(44.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) = \(44.0 \times 0.100 \times 2~\mathrm{mmol}\) Moles of SO42- in \(44.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) = \(44.0 \times 0.100~\mathrm{mmol}\) Moles of K+ in \(25.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{KCl}\) = \(25.0 \times 0.150~\mathrm{mmol}\) Moles of Cl- in \(25.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{KCl}\) = \(25.0 \times 0.150~\mathrm{mmol}\) (Tag steps 2-4 are the same as the steps in example a, but with the calculated moles for this case.) After calculating the values for each ion, we obtain: Concentration of Na+ = \(0.127~\mathrm{M}\) Concentration of SO42- = \(0.0636~\mathrm{M}\) Concentration of K+ = \(0.0469~\mathrm{M}\) Concentration of Cl- = \(0.0469~\mathrm{M}\) (c) Mixing \(3.60 \mathrm{~g}\) of \(KCl\) in \(75.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{CaCl}_{2}\) solution (Tag steps 1-4 are the same as the steps in example a, but with the calculated moles for this case.) After calculating the values for each ion, we obtain: Concentration of K+ = \(0.538~\mathrm{M}\) Concentration of Cl- = \(1.08~\mathrm{M}\) Concentration of Ca2+ = \(0.214~\mathrm{M}\)

Key Concepts

Molarity

Molarity is a key term when discussing solution concentration, as it defines the number of moles of a solute that is present in one liter of solution. It's an easy way to express how concentrated a solution is.
To calculate molarity, simply divide the number of moles of the solute by the volume of the solution in liters. For example, if you have a solution with 0.170 moles of NaOH in a 1-liter solution, its molarity is 0.170 M. Molarity is often denoted as "M" followed by the solute, such as "0.170 M NaOH".
Understanding molarity helps us determine how many particles of a particular solute are dispersed in a solution, which is crucial for predicting how the solution will behave chemically.

Ion Concentration

Ion concentration refers to the amount of individual ions in a solution. This helps us find out how much of each ion is present, which can affect the solution's properties.
To determine ion concentrations, consider the compound being dissolved. For example, NaOH dissociates into Na⁺ and OH^- ions in solution. If the concentration of NaOH is known, it directly corresponds to the concentration of Na⁺ and OH^-, as each unit of NaOH gives one Na⁺ and one OH^-.
This concept becomes more intricate when dealing with compounds that produce multiple ions, such as Na₂SO₄, which yields two Na⁺ ions and one SO₄^2- ion for every unit dissolved.

Chemical Solutions

Chemical solutions are comprised of solutes dissolved in solvents. In most cases, the solvent is a liquid, and the solute is the substance being dissolved, which can be a solid, liquid, or gas.
These solutions are uniform mixtures, meaning that the composition is consistent throughout. For example, a saltwater solution is a mixture where salt is the solute and water is the solvent, forming a homogeneous liquid mixture.
Understanding the nature of chemical solutions helps us with tasks such as calculating ion concentrations and molarities. Importantly, solutions play a vital role in various chemical reactions and processes both in nature and industrial applications.

Mixing Solutions

Mixing solutions involves combining two or more solutions to form a single mixture. When solutions are mixed, one key assumption we make is that the volumes are additive, meaning you just add the volumes of the solutions together to predict the final volume.
After mixing, the concentration of ions changes because they are being dispersed throughout a new volume. To find the new concentrations, we calculate how many moles of ions are present initially and then divide by the total volume after mixing.
This concept is crucial in chemistry, particularly in neutralization reactions or when preparing solutions with specific concentrations. Understanding how to accurately mix solutions allows chemists to predict reactions' outcomes reliably.