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

Expert verified
The final molarity of KBr in the solution is approximately \(1.42\mathrm{M}\).

Step by step solution

01

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: \(moles_{1} = volume_{1} \times molarity_{1}\) Moles of KBr in the second solution: \(moles_{2} = volume_{2} \times molarity_{2}\)
02

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: \(moles_{total} = moles_{1} + moles_{2}\)
03

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: \(molarity_{final} = \frac{moles_{total}}{volume_{final}}\) Now, we can plug in the numbers and solve for the molarity of KBr in the final solution.
04

Plug in the values and solve

Moles of KBr in the first solution: \(moles_{1}=(35.0\times10^{-3}\mathrm{L})\times(1.00\mathrm{M})= 0.035\mathrm{mol}\) Moles of KBr in the second solution: \(moles_{2}=(60.0\times10^{-3}\mathrm{L})\times(0.600\mathrm{M})= 0.036\mathrm{mol}\) Total moles of KBr: \(moles_{total} = 0.035 \mathrm{mol} + 0.036 \mathrm{mol} = 0.071\mathrm{mol}\) Molarity of KBr in the final solution: \(molarity_{final} = \frac{0.071\mathrm{mol}}{50.0\times10^{-3}\mathrm{L}} = 1.42\mathrm{M}\) The final molarity of KBr in the solution is approximately \(1.42\mathrm{M}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:
\[ Molarity (M) = \frac{Number\ of\ moles\ of\ solute}{Volume\ of\ solution\ in\ liters} \]
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:\[ \begin{align*} M_1 &= \text{initial molarity of the solution} \ V_1 &= \text{initial volume of the solution} \ M_2 &= \text{final molarity of the diluted solution} \ V_2 &= \text{final volume of the diluted solution} \end{align*} \]
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.

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Most popular questions from this chapter

The average concentration of bromide ion in seawater is \(65 \mathrm{mg}\) of bromide ion per \(\mathrm{kg}\) of seawater. What is the molarity of the bromide ion if the density of the seawater is \(1.025 \mathrm{~g} / \mathrm{mL} ?\)

(a) How many milliliters of a stock solution of \(6.0 \mathrm{M} \mathrm{HNO}_{3}\) would you have to use to prepare \(110 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{HNO}_{3} ?\) (b) If you dilute \(10.0 \mathrm{~mL}\) of the stock solution to a final volume of \(0.250 \mathrm{~L}\), what will be the concentration of the diluted solution?

Label each of the following substances as an acid, base, salt, or none of the above. Indicate whether the substance exists in aqueous solution entirely in molecular form, entirely as ions, or as a mixture of molecules and ions. (a) HF, (b) acetonitrile, \(\mathrm{CH}_{3} \mathrm{CN},\) (c) \(\mathrm{NaClO}_{4},\) (d) \(\mathrm{Ba}(\mathrm{OH})_{2}\).

Explain the following observations: (a) \(\mathrm{NH}_{3}\) contains no \(\mathrm{OH}^{-}\) ions, and yet its aqueous solutions are basic; (b) HF is called a weak acid, and yet it is very reactive; (c) although sulfuric acid is a strong electrolyte, an aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) contains more \(\mathrm{HSO}_{4}^{-}\) ions than \(\mathrm{SO}_{4}^{2-}\) ions.

Suppose you have a solution that might contain any or all of the following cations: \(\mathrm{Ni}^{2+}, \mathrm{Ag}^{+}, \mathrm{Sr}^{2+},\) and \(\mathrm{Mn}^{2+}\). Addition of HCl solution causes a precipitate to form. After filtering off the precipitate, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to the resulting solution and another precipitate forms. This is filtered off, and a solution of \(\mathrm{NaOH}\) is added to the resulting solution. No precipitate is observed. Which ions are present in each of the precipitates? Which of the four ions listed above must be absent from the original solution?

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