Question

Calculate the sodium ion concentration when \(70.0 \mathrm{mL}\) of 3.0 \(M\) sodium carbonate is added to \(30.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate.

Short answer

The sodium ion concentration in the final solution after mixing \(70.0 \mathrm{mL}\) of 3.0 \(M\) sodium carbonate and \(30.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate is \(4.50 \mathrm{M}\).

Step-by-step solution

List the given information and find the moles of sodium ions in each solution.

We are given the following information: - Sodium carbonate solution: \(70.0 \mathrm{mL}\) volume and \(3.0 \mathrm{M}\) concentration (note that one mole of sodium carbonate \(\mathrm{Na_2CO_3}\) contains two moles of sodium ions \(\mathrm{Na^+}\)) - Sodium bicarbonate solution: \(30.0 \mathrm{mL}\) volume and \(1.0 \mathrm{M}\) concentration (one mole of sodium bicarbonate \(\mathrm{NaHCO_3}\) contains one mole of sodium ions \(\mathrm{Na^+}\)) To find the moles of sodium ions in each solution we use the formula: moles = volume × concentration. Note we need to multiply with the number of ions in each molecule for Sodium carbonate. Moles of sodium ions in sodium carbonate solution: \(2 \times (70.0 \mathrm{mL} \times 3.0 \mathrm{M}) = 2 \times (210.0 \mathrm{mmol}) = 420.0 \mathrm{mmol}\) Moles of sodium ions in sodium bicarbonate solution: \(30.0 \mathrm{mL} \times 1.0 \mathrm{M} = 30.0 \mathrm{mmol}\)

Find the total volume of the mixed solution.

Next, we need to find the total volume of the final solution. To do this, simply add the volumes of each solution together: Total volume = volume of sodium carbonate + volume of sodium bicarbonate Total volume = \(70.0 \mathrm{mL} + 30.0 \mathrm{mL} = 100.0 \mathrm{mL}\)

Calculate the total moles of sodium ions in the final solution.

Now, we need to find the total moles of sodium ions in the final solution. To do this, add the moles of sodium ions from each solution: Total moles = moles of sodium ions in sodium carbonate + moles of sodium ions in sodium bicarbonate Total moles = \(420.0 \mathrm{mmol} + 30.0 \mathrm{mmol} = 450.0 \mathrm{mmol}\)

Calculate the sodium ion concentration in the final solution.

Finally, we can find the sodium ion concentration in the final solution by dividing the total moles of sodium ions by the total volume of the solution: Sodium ion concentration = \(\frac{\text{Total moles}}{\text{Total volume}}\) Sodium ion concentration = \(\frac{450.0 \mathrm{mmol}}{100.0 \mathrm{mL}} = 4.50 \mathrm{M}\) Therefore, the sodium ion concentration in the final solution is \(4.50 \mathrm{M}\).

Key concepts

Molarity Calculation

Molarity is a way to express the concentration of a solution, and it's an essential concept in chemistry. It tells us how many moles of a solute are present in one liter (or one thousand milliliters) of a solution. To calculate molarity, the formula used is:

• Molarity (M) = moles of solute / liters of solution

In the problem at hand, the exercise asks for the concentration of sodium ions in the final mixed solution. First, we determine the number of moles of sodium ions in each individual solution by multiplying the volume of the solution (in liters) by its molarity. Remember, for sodium carbonate, each molecule releases two sodium ions, so the moles calculated must be multiplied by two. Thus, for each sodium compound:

• Sodium carbonate, \( ext{Moles} = 2 \times (0.070 \, L) \times (3.0 \, M) = 420.0 \, mmol \)
• Sodium bicarbonate, \( ext{Moles} = (0.030 \, L) \times (1.0 \, M) = 30.0 \, mmol \)

This gives us a straightforward way to evaluate the molarity of sodium ions once the solution is mixed.

Solution Mixing

When mixing two solutions, understanding how they integrate into one complete solution is vital. Solution mixing involves combining volumes of two solutions to form a new one. In this problem, two different sodium salt solutions are mixed together, hence we need to add their respective volumes to find the total volume.The formula to find the total volume after mixing is simple:

• Total volume = Volume of solution 1 + Volume of solution 2

In this problem:

• From sodium carbonate, \( 70.0 \, ext{mL} \)
• From sodium bicarbonate, \( 30.0 \, ext{mL} \)
• Total Volume = \( 70.0 \, ext{mL} + 30.0 \, ext{mL} = 100.0 \, ext{mL} \)

Now that we know the total volume of the new solution, this can be utilized to calculate the final concentration of solutes in that solution. This calculation is crucial for ensuring the accuracy of the result in chemistry problems involving solution mixtures.

Stoichiometry

Stoichiometry is the part of chemistry that focuses on quantifying the substances consumed and produced in chemical reactions. It is a powerful tool for solving chemical equations, allowing us to predict the amount of reactants needed or products produced. In the context of this exercise, stoichiometry helps determine the number of sodium ions provided by each compound.When dealing with stoichiometry, always consider the relationship between different chemical species, often using balanced chemical equations to predict these ratios. Remember:

• A mole is a unit that provides a way to count particles such as atoms, ions, or molecules, and it's fundamental in stoichiometry.
• In sodium carbonate (\( ext{Na}_2 ext{CO}_3 \)), every mole gives two moles of sodium ions because there are two sodium atoms in its formula.
• Similarly, for sodium bicarbonate (\( ext{NaHCO}_3 \)), each mole corresponds to one mole of sodium ions as it provides one sodium atom per molecule.

By understanding the stoichiometric relationships of sodium ions in different compounds, we can calculate their moles accurately and, in turn, find the final concentration when solutions are mixed.