Question

A $35.0-\mathrm{mL}$ sample of $1.00\mathrm{M}\mathrm{KBr}$ and a $60.0-\mathrm{mL}$ sample of $0.600\mathrm{M}\mathrm{KBr}$ are mixed. The solution is then heated to evaporate water until the total volume is $50.0\mathrm{mL}$. What is the molarity of the KBr in the final solution?

Short answer

The final molarity of KBr in the solution is approximately $1.42\mathrm{M}$.

Step-by-step solution

Calculate the initial moles of KBr in each solution

We can find the moles of KBr in each solution by multiplying their volume (in L) with their respective molarities (M). Moles of KBr in the first solution: $mole{s}_1=volum{e}_1\times molarit{y}_1$ Moles of KBr in the second solution: $mole{s}_2=volum{e}_2\times molarit{y}_2$

Find the total moles of KBr in the mixture

We can obtain the total moles of KBr in the mixture by adding the moles from both solutions. Total moles of KBr: $mole{s}_{total}=mole{s}_1+mole{s}_2$

Calculate the molarity of KBr in the final solution

Since the number of moles of KBr doesn't change during evaporation, we can calculate the molarity of KBr in the final solution using the total moles of KBr and the final volume: Molarity of KBr in the final solution: $molarit{y}_{final}=\frac{mole{s}_{total}}{volum{e}_{final}}$ Now, we can plug in the numbers and solve for the molarity of KBr in the final solution.

Plug in the values and solve

Moles of KBr in the first solution: $mole{s}_1=(35.0\times {10}^{-3}\mathrm{L})\times (1.00\mathrm{M})=0.035\mathrm{mol}$ Moles of KBr in the second solution: $mole{s}_2=(60.0\times {10}^{-3}\mathrm{L})\times (0.600\mathrm{M})=0.036\mathrm{mol}$ Total moles of KBr: $mole{s}_{total}=0.035\mathrm{mol}+0.036\mathrm{mol}=0.071\mathrm{mol}$ Molarity of KBr in the final solution: $molarit{y}_{final}=\frac{0.071\mathrm{mol}}{50.0\times {10}^{-3}\mathrm{L}}=1.42\mathrm{M}$ The final molarity of KBr in the solution is approximately $1.42\mathrm{M}$.

Key concepts

Molarity and Concentration

Understanding molarity is fundamental to grasping the concepts of solution chemistry. Molarity, denoted by the symbol 'M', is a measure of concentration in chemistry that represents the number of moles of a solute that is dissolved in one liter of solution. It tells us how concentrated a solution is with regards to a particular substance.

Let's consider the term 'solute'—it's the substance that is dissolved in a liquid (known as the solvent) to make a solution. For instance, if we dissolve salt (NaCl) in water (H₂O), salt is the solute, and water is the solvent. The resulting saline water is the solution.

To calculate molarity, we use the formula:
$$Molarity(M)=\frac{Numberofmolesofsolute}{Volumeofsolutioninliters}$$
So, if you dissolve 1 mole of salt in 1 liter of water, the molarity of that salt solution is 1 M. If you have more or fewer moles, or a different volume, the molarity will change accordingly. Understanding molarity allows students to perform various calculations, such as determining the amount of a substance required to create a desired concentration of solution.

Stoichiometry of Solutions

Stoichiometry refers to the calculation of reactants and products in chemical reactions. In the context of solutions, stoichiometry involves the quantitative relationships between the amounts of solutes in a given volume of solution.

Using stoichiometry, we can relate the molarity of solutions to each other and to the amounts of reactants and products in chemical reactions. All this is done through moles, as it's a reliable way to measure substances due to Avogadro's number, which links the micro world of atoms and molecules to a macro scale we can work with.

To perform stoichiometric calculations for solutions:

• Determine the molarity of the solution(s) involved.
• Convert between moles, mass, and volume as required.
• Apply the mole ratios from the balanced chemical equation if a reaction is involved.

Having a balanced chemical equation is important, as it shows the ratio of reactants to products, and this ratio is fundamental in stoichiometric calculations. This allows us to understand big picture concepts like how much product we'll make from a given amount of reactant or how much of one reactant we'll need to fully react with another.

Solution Dilution Calculations

When we dilute a solution, we're simply adding more solvent to a smaller volume of a more concentrated solution to achieve a lesser concentration. This is very important in laboratory and clinical settings, where precise concentrations are crucial.

To calculate dilutions, there's a simple mathematical relationship expressed by the equation:
$$M_1\times V_1=M_2\times V_2$$
Where:$$\begin{array}{rlrlr}{M}_1& =\text{initial molarity of the solution}{V}_1& =\text{initial volume of the solution}{M}_2& =\text{final molarity of the diluted solution}{V}_2& =\text{final volume of the diluted solution}\end{array}$$
So, if you know any three parts of the equation, you can calculate the fourth. This is handy when you need to create a solution of a certain molarity from a stock solution that is more concentrated. Dilution doesn't alter the amount of solute present; it just spreads it out more thinly in the increased volume of solvent.