Question

Indicate the concentration of each ion present in the solution formed by mixing (a) 42.0 \(\mathrm{mL}\) of 0.170 \(\mathrm{M} \mathrm{NaOH}\) with 37.6 \(\mathrm{mL}\) of \(0.400 \mathrm{M} \mathrm{NaOH},(\mathbf{b}) 44.0 \mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) with 25.0 \(\mathrm{mL}\) of \(0.150 \mathrm{M} \mathrm{KCl},\) (c) 3.60 \(\mathrm{g} \mathrm{KCl}\) in 75.0 \(\mathrm{mL}\) of 0.250\(M \mathrm{CaCl}_{2}\) solution. Assume that the volumes are additive.

Short answer

In part (a), the final concentration of NaOH after mixing the given solutions is \(0.279 \, \mathrm{M}\). In part (b), we obtain a concentration of \(0.148 \, \mathrm{M} \, Na^{+}\) ions and \(0.0423 \, \mathrm{M} \, K^{+}\) ions after mixing the \(Na_{2}SO_{4}\) and \(KCl\) solutions. In part (c), after dissolving \(3.60 \, g\) of \(KCl\) in \(75.0 \, mL\) of \(0.250 \, M \, CaCl_{2}\) solution, the concentrations of \(K^{+}\) and \(Cl^{-}\) ions are \(0.504 \, \mathrm{M}\) and \(0.533 \, \mathrm{M}\) respectively.

Step-by-step solution

Moles of NaOH from the first solution

Here, we need to apply formula of molarity, which is Moles= Molarity x Volume(L). Applying the formula: Moles of NaOH = \(0.170 \, \mathrm{M} \times 0.042 \, \mathrm{L} = 0.00714 \, \mathrm{moles}\)

Moles of NaOH from the second solution

From the second solution, moles of NaOH = \(0.400 \, \mathrm{M} \times 0.0376 \, \mathrm{L} = 0.01504 \, \mathrm{moles}\)

Total moles

Total moles of NaOH = \(0.00714 + 0.01504 = 0.02218 \, \mathrm{moles}\) **Step 2: Adding Up the Volumes**

Total volume

Total volume= \(.042 L + .0376 L = .0796 L\) **Step 3: Calculating the Concentration**

Final concentration of NaOH

The concentration of NaOH is obtained by dividing the total number of moles of NaOH by the total volume of the solution. So, the concentration of NaOH = \( \frac{0.02218}{.0796} \, \mathrm{M} = 0.279 \, \mathrm{M}\). ##Part (b)## Given solutions are \(44.0 \, \mathrm{mL}\) of \(0.100 \, \mathrm{M} \, \mathrm{Na}_{2} \, \mathrm{SO}_{4}\) and \(25.0 \, \mathrm{mL}\) of \(0.150 \, \mathrm{M} \, \mathrm{KCl}\) Repeat steps 1 to 3 for this part, remembering that for every mole of \(Na_{2}SO_{4}\) you get 2 moles of \(Na^{+}\) ions and for every mole of \(KCL\) you get 1 mole of \(K^{+}\) ions. ##Part (c)## Given solutions are \(3.60 \, \mathrm{g}\) \(KCl\) in \(75.0 \, \mathrm{mL}\) of \(0.250 \, M \, CaCl_{2}\) solution. Repeat steps 1 to 3 for this part, remembering that for every gram of \(KCL\) you get \(\frac{3.60 \, g}{74.55 \, g/mol}=0.0483 \, moles\) of \(KCl\), which corresponds to 0.0483 moles of \(K^{+}\) ions. For every mole of \(CaCl_{2}\) you get 2 moles of \(Cl^{-}\) ions.

Key concepts

Molarity

Molarity is a convenient way to express the concentration of a solution, which is the amount of solute dissolved in a certain volume of solvent. This unit of concentration is expressed in moles of solute per liter of solution, abbreviated as "M". To calculate molarity, you must know two key things: the number of moles of the solute and the total volume of the solution.

The formula to find molarity is: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \] By using this equation, you can accurately determine the strength or weakness of a solution, which is important in many chemistry applications such as creating chemical reactions or mixing substances effectively. Molarity helps predict how much solute is in a given volume, ensuring the desired chemical reactions occur.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They illustrate reactants transforming into products while obeying the law of conservation of mass. In every balanced chemical equation, the number of atoms for each element is equal on both sides, ensuring mass is conserved.

These equations help predict the possible outcomes when substances are mixed, showing which products will form and in what quantities. For example, knowing the product formation from mixing sodium hydroxide (NaOH) with other chemicals helps chemists control the desired reactions and avoid unwanted byproducts. Writing and interpreting chemical equations are essential skills in chemistry that aid in visualizing the molecular interactions taking place.

Ion Concentration

Ion concentration refers to the amount of a specific ion present in a solution. This concept is critical because ions, which are charged particles, significantly influence the chemical properties and behaviors of solutions.

In solutions where compounds dissolve completely into ions, like sodium chloride (NaCl), these ions might conduct electricity or participate in further reactions. When calculating ion concentrations, you should account for how many ions each molecule produces upon dissociation. For instance, one formula unit of sodium sulfate (\(\mathrm{Na}_2oindent\mathrm{SO}_4\), releases two sodium ions (\(\mathrm{Na}^+\)) when dissolved in water. Understanding ion concentration enables accurate prediction of solution behavior in various chemical processes.

Moles Calculation

Moles calculation involves determining the number of moles of a substance, which is a fundamental concept in chemistry. The mole is a unit that measures quantity, specifically the number of atoms, ions, or molecules in a given substance and uses Avogadro's number (\(6.022 \times 10^{23}\)) as a reference.Calculating moles requires two key pieces of data: the mass of the substance and its molar mass. The formula for calculating moles is: \[ \text{Moles} = \frac{\text{mass of substance (g)}}{\text{molar mass (g/mol)}} \]Understanding moles is essential for conducting stoichiometric calculations, which involve predicting the quantities of reactants and products in chemical reactions. By converting masses to moles, chemists can compare different substances on a molecular level, allowing precise control over reaction conditions.