Question

(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{MNH}_{3} ?\) (b) If you take a \(10.0-\mathrm{mL}\) portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L},\) what will be the concentration of the final solution?

Short answer

(a) To make a \(1000.0\,\mathrm{mL}\) solution of \(0.250\,\mathrm{M}\,\mathrm{NH}_{3}\), you will need approximately \(16.89\,\mathrm{mL}\) of the \(14.8\,\mathrm{M}\,\mathrm{NH}_{3}\) stock solution. (b) The concentration of the final solution after diluting \(10.0\,\mathrm{mL}\) of the stock solution to \(0.500\,\mathrm{L}\) is approximately \(0.296\,\mathrm{M}\).

Step-by-step solution

Identify given values and setup the formula

Given: \(c_{1} = 14.8\,\mathrm{M}\), \(c_{2} = 0.250\,\mathrm{M}\), and \(v_{2} = 1000.0\,\mathrm{mL}\). We need to find \(v_{1}\). Formula: \(c_{1}v_{1} = c_{2}v_{2}\).

Solve for \(v_{1}\)

Rearrange the formula to solve for \(v_{1}\): \(v_{1} = \frac{c_{2}v_{2}}{c_{1}}\) Now, plug in the values: \(v_{1} = \frac{(0.250\,\mathrm{M})(1000.0\,\mathrm{mL})}{14.8\,\mathrm{M}}\)

Calculate \(v_{1}\)

Perform the calculation: \(v_{1} \approx 16.89\,\mathrm{mL}\) So, approximately \(16.89\,\mathrm{mL}\) of the stock solution is needed to make \(1000.0\,\mathrm{mL}\) of \(0.250\,\mathrm{M}\,\mathrm{NH}_{3}\). #b) Calculate the concentration of the final solution after diluting the 10.0 mL stock solution#

Identify given values and setup the formula

Given: \(c_{1} = 14.8\,\mathrm{M}\), \(v_{1} = 10.0\,\mathrm{mL}\), and \(v_{2} = 0.500\,\mathrm{L} (= 500.0\,\mathrm{mL})\). We need to find \(c_{2}\). Formula: \(c_{1}v_{1} = c_{2}v_{2}\).

Solve for \(c_{2}\)

Rearrange the formula to solve for \(c_{2}\): \(c_{2} = \frac{c_{1}v_{1}}{v_{2}}\) Now, plug in the values: \(c_{2} = \frac{(14.8\,\mathrm{M})(10.0\,\mathrm{mL})}{500.0\,\mathrm{mL}}\)

Calculate \(c_{2}\)

Perform the calculation: \(c_{2} \approx 0.296\,\mathrm{M}\) So, the concentration of the final solution after diluting the \(10.0\,\mathrm{mL}\) stock solution to \(0.500\,\mathrm{L}\) is approximately \(0.296\,\mathrm{M}\).

Key concepts

Molarity

Molarity is a key concept in chemistry that measures the concentration of a solute in a solution. It is defined as the number of moles of a solute per liter of solution, often expressed in moles per liter (\(\text{mol/L}\)). Understanding molarity is important for preparing solutions with precise concentrations. Here's a simple way to think about it:

• A higher molarity means a more concentrated solution, with more solute particles in a given volume of solvent.
• A lower molarity indicates a more dilute solution, with fewer solute particles.

Imagine you're preparing lemonade. The lemon taste is stronger with more syrup in the same amount of water, similar to a solution with high molarity. Conversely, adding more water with the same amount of syrup produces a weaker flavor, much like a solution with low molarity. Molarity is crucial for describing how strong or weak a chemical solution is.

Solution Concentration

Solution concentration is a measure of how much solute is present in a solution relative to the amount of solvent. It's essential for understanding how to create solutions for various applications in chemistry, from laboratory experiments to industrial processes.

• Concentration indicates how rich or lean a solution is in terms of solute content.
• Common units of concentration include molarity (mol/L), molality (mol/kg), and percent composition (weight/weight), although molarity is the most frequently used in dilution calculations.

In dilution problems, you might know the desired concentration and volume for a final solution. Here, you use the initial stock solution's concentration to determine how much of it is needed. In everyday terms, it's like knowing how much concentrate you have and working out how much water to add to get the desired drink strength. This approach ensures solutions meet specific requirements for reactions, analysis, or consumption.

Chemistry Formulas

Chemistry formulas are the backbone of problem-solving in chemistry. They provide a systematic approach to calculate unknown quantities using known values. A key formula used in dilution calculations is:\[ c_{1}v_{1} = c_{2}v_{2} \]In this equation:

• \(\,c_{1}\,\) is the concentration of the initial solution (stock solution).
• \(\,v_{1}\,\) is the volume of the initial solution you will use.
• \(\,c_{2}\,\) is the concentration of the final solution.
• \(\,v_{2}\,\) is the volume of the final solution.

These symbols and the equations make it straightforward to rearrange terms and calculate any unknown parameter as long as the other three are known. Whether you're determining how much of a concentrated liquid is needed to make a weaker solution or finding the final concentration after dilution, this formula simplifies the process neatly. Chemistry formulas, hence, enable us to accurately translate real-world problems into computable solutions.