Practise the calculations involved in dilutions (answer in each case to four decimal places). (a) If you added sodium chloride solution \((1.00 \mathrm{~mL} ; 0.4 \mathrm{M}\) ) to a volumetric flask \((10.00 \mathrm{~mL})\) and made up to the mark with water, what would be the concentration of the diluted solution? (b) Calculate the volume a solution of copper (II) sulphate \((0.1 \mathrm{M})\) required to produce \(500.00\) \(\mathrm{mL}\) of \(0.02 \mathrm{M}\) solution. (c) What would be the concentration of a solution of \(\mathrm{Fe}^{2+}\) ions if \(10.00 \mathrm{~mL}\) is diluted to \(250.00 \mathrm{~mL}\) to give a concentration of \(0.001 \mathrm{M} ?\) (d) You are provided with a solution of \(\mathrm{KMnO}_{4}(0.02 \mathrm{M})\); to what volume must \(5.00 \mathrm{~mL}\) of this solution be diluted to give a concentration of \(0.001 \mathrm{M}\) ?

Step by step solution

01

Solve for (a)

Use the dilution formula \(M_1V_1 = M_2V_2\). Here, \(M_1\) is 0.4 M, \(V_1\) is 1 mL, and \(V_2\) is 10 mL. Solve for \(M_2\), which gives the concentration of the diluted solution: \(M_2 = (M_1V_1) / V_2 = (0.4M * 1mL) / 10mL = 0.04M\).

02

Solve for (b)

Again use the dilution formula. Here, \(M_1\) is 0.1 M, \(M_2\) is 0.02 M, and \(V_2\) is 500 mL. Solve for \(V_1\), the initial volume needed: \(V_1 = (M_2V_2) / M_1 = (0.02M * 500mL) / 0.1M = 100mL\).

03

Solve for (c)

Using the dilution formula, you want to find \(M_1\). Here, \(M_2\) is 0.001 M, \(V_1\) is 10 mL, and \(V_2\) is 250 mL: \(M_1 = (M_2V_2) / V_1 = (0.001M * 250mL) / 10mL = 0.025M\).

04

Solve for (d)

Use the dilution formula to find \(V_2\). Here, \(M_1\) is 0.02 M, \(M_2\) is 0.001 M, and \(V_1\) is 5 mL: \(V_2 = (M_1V_1) / M_2 = (0.02M * 5mL) / 0.001M = 100mL\).

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Key Concepts

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

Molarity and Dilution

Understanding the relationship between molarity and dilution is essential in chemistry for preparing solutions of precise concentrations. Molarity, represented as M, is the measure of concentration indicating moles of a solute per liter of solution. When you dilute a solution, you essentially add more solvent without adding more solute, which decreases the solution's concentration.

Let's consider the given exercise, which involves diluting a sodium chloride solution. The dilution formula, which is a cornerstone in these calculations, is given by: \( M_1V_1 = M_2V_2 \). Here, \( M_1 \) and \( M_2 \) are the molarities of the concentrated and diluted solutions, respectively, while \( V_1 \) and \( V_2 \) denote their volumes. By manipulating this equation, you can solve for any unknown in dilution problems.

Chemical Solution Concentration

Concentration in chemistry dictates how much solute is present in a given amount of solvent or solution. It’s fundamental for predicting how substances will react and the yield of products from these reactions. Concentrations can be expressed in several ways, with molarity being one widely used unit. Other units include molality, normality, and mass percentage.

In the context of the problems mentioned, molarity is used to find the concentration after diluting a solution of known initial concentration and volume. It's important to remember that the amount of solute remains the same before and after dilution; it's the volume that changes, thereby affecting concentration. Practicing these calculations, as seen in the provided steps, will strengthen your understanding of how concentrations vary with volume changes.

Stoichiometry in Solutions

Stoichiometry extends beyond reactions to also involve solutions and their components. It allows chemists to calculate volumes or concentrations of solutes in a solution following a dilution or reaction. Stoichiometry in solutions is deeply intertwined with molarity since it expresses the quantitative relationship between reactants and products in a homogeneous mixture.

In the questions considered, stoichiometry highlights how the concentration of solutions change when diluted. This process does not change the number of moles of solute present, just the total volume of the solution, and thus the concentration. Mastery of stoichiometry is crucial for making solutions with precise molarities, which is vital for laboratory work, industrial applications, and even pharmaceutical preparations.