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\(M\) sodium bicarbonate.

Short answer

The sodium ion concentration in the solution is 4.5 M.

Step-by-step solution

Convert volumes to liters

Since the volumes of solutions are given in milliliters, we need to convert them to liters for easier calculations. To convert from milliliters to liters, divide by 1000. - 70.0 mL = 0.070 L for sodium carbonate solution - 30.0 mL = 0.030 L for sodium bicarbonate solution

Calculate moles of sodium ions in sodium carbonate solution

To calculate the moles of sodium ions present in sodium carbonate solution, we use the formula: moles = molarity × volume. In sodium carbonate (Na2CO3), there are 2 sodium ions per molecule. Therefore, the moles of sodium ion can be calculated as: Moles in Na2CO3 = Molarity of sodium carbonate × Volume of sodium carbonate × Number of sodium ions per molecule Moles of Na in Na2CO3 = 3.0 mol/L × 0.070 L × 2 = 0.42 mol

Calculate moles of sodium ions in sodium bicarbonate solution

Similarly for sodium bicarbonate (NaHCO3), there is 1 sodium ion per molecule. Calculate the moles of sodium ions by using the formula: moles = molarity × volume. Moles of Na in NaHCO3 = Molarity of sodium bicarbonate × Volume of sodium bicarbonate × Number of sodium ions per molecule Moles of Na in NaHCO3 = 1.0 mol/L × 0.030 L × 1 = 0.03 mol

Calculate the total moles of sodium ions

Add the moles of sodium ions from both sodium carbonate and sodium bicarbonate solutions to find the total moles of sodium ions in the mixture. Total moles of Na = Moles of Na in Na2CO3 + Moles of Na in NaHCO3 Total moles of Na = 0.42 mol + 0.03 mol = 0.45 mol

Calculate the total volume of the mixture

To find the total volume of the mixture, add the volumes of sodium carbonate and sodium bicarbonate solutions together. Total volume = Volume of sodium carbonate solution + Volume of sodium bicarbonate solution Total volume = 0.070 L + 0.030 L = 0.100 L

Calculate the concentration of sodium ions

Finally, divide the total moles of sodium ions by the total volume of the mixture to find the concentration of sodium ions in the mixture. Concentration of Na = Total moles of Na / Total volume Concentration of Na = 0.45 mol / 0.100 L = 4.5 M The sodium ion concentration in the solution is 4.5 M.

Key concepts

Molarity

Molarity is a crucial concept in chemistry that refers to the concentration of a solute in a solution. It is defined as the number of moles of a solute divided by the volume of the solution in liters. The formula for molarity is:\[ \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}} \]Understanding molarity helps in calculating how much solute is present in a solution, which is essential for reacting mixtures accurately.

• Molarity equals moles divided by liters.
• Used to express concentration.
• Helps in preparing solutions and reacting chemicals accurately.

When dealing with problems like calculating sodium ion concentrations, molarity helps in determining how much solute, like sodium carbonate or sodium bicarbonate, is present in each solution. Converting volumes from milliliters to liters ensures calculations are accurate within this context.

Sodium Carbonate

Sodium carbonate, often known by its chemical formula \(\text{Na}_2\text{CO}_3\), is a common chemical compound used in various household and industrial applications. Each molecule contains two sodium ions, which makes it particularly interesting when calculating ion concentrations.
Sodium carbonate is fully soluble in water, dissociating into its respective ions. In calculations involving sodium carbonate, remember that each mole of \(\text{Na}_2\text{CO}_3\) contributes two moles of sodium ions to the solution. This factor is crucial when determining the overall sodium concentration from sodium carbonate solutions.

• Ironically known as "soda ash."
• Each molecule provides two sodium ions.
• Used in glass manufacturing and water softening.

In chemical calculations involving sodium carbonate, accounting for the two sodium ions per molecule is key to arriving at correct concentrations.

Sodium Bicarbonate

Sodium bicarbonate, known chemically as \(\text{NaHCO}_3\), is another compound often used in baking and science experiments. Unlike sodium carbonate, each \(\text{NaHCO}_3\) molecule provides one sodium ion upon dissolution.
This makes sodium bicarbonate valuable for calculations where a precise sodium ion concentration is needed. When added to solutions, it maintains a lower contribution of sodium ions compared to sodium carbonate for the same molarity.

• Commonly referred to as "baking soda."
• Each molecule provides one sodium ion.
• Used in cooking, cleaning, and neutralizing acids.

In any problem involving mixing solutions, knowing the number of ions each compound contributes helps in achieving the desired concentration.

Solution Mixing

Solution mixing involves combining two or more solutions to achieve a target concentration or to create a reaction. The process requires understanding the volumes and molarities of the involved solutions.
For example, when mixing sodium carbonate and sodium bicarbonate solutions, the total volume is the sum of the individual volumes.

• Mixing involves calculating combined volume.
• Key to achieving equilibrium and desired concentrations.
• Often needed in chemical reactions and laboratory preparations.

Proper solution mixing ensures that the sodium ion concentration is consistent throughout the mixture, leading to accurate results when further chemical interactions are intended.

Chemical Calculations

Chemical calculations are essential for accurately preparing mixtures and predicting the results of reactions. They encompass converting between units, such as milliliters to liters, calculating moles of substances, and determining concentrations.
When performing chemical calculations for sodium ion concentrations, add up the moles of ions from each solution and divide by total volume for the mixture.

• Essential for unit conversions and determining concentration.
• Helps maintain accuracy in experiments and reactions.
• Relies on using correct formulas and understanding of reactants.

Mastering chemical calculations is vital for exploring chemistry effectively, ensuring each step and computation aligns perfectly with theoretical expectations.