Question

Describe how you would prepare each of the following aqueous solutions: (a) $1.50\mathrm{L}$ of $0.110\mathrm{M}{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ solution, starting with solid ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4;$ (b) $225\mathrm{g}$ of a solution that is $0.65\mathrm{m}$ in ${\mathrm{Na}}_2{\mathrm{CO}}_3,$ starting with the solid solute; (c) $1.20\mathrm{L}$ of a solution that is $15.0\mathrm{\%}\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ by mass (the density of the solution is $1.16\mathrm{g}/\mathrm{mL}$ ), starting with solid solute; (d) a $0.50M$ solution of HCl that would just neutralize $5.5\mathrm{g}$ of $\mathrm{Ba}(\mathrm{OH}{)}_2$ starting with $6.0\mathrm{M}\mathrm{HCl}$.

Short answer

$a)$: To prepare 1.50 L of a 0.110 M solution of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$, weigh 21.80 g of the solid and dissolve it in water, then add water until the total volume of the solution is 1.50 L. $b)$: To prepare 225 g of a solution with 0.65 m ${\mathrm{Na}}_2{\mathrm{CO}}_3$, weigh 15.5 g of solid and dissolve it in water, then add water until the total mass of the solution is 225 g. $c)$: To prepare 1.20 L of a solution with 15.0% $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ by mass, weigh 208.8 g of solid solute and dissolve it in water, then add water until the total mass of the solution is 1392 g. $d)$: To prepare a 0.50 M HCl solution that would neutralize 5.5 g of $\mathrm{Ba}(\mathrm{OH}{)}_2$, take 10.7 mL of 6 M HCl, dilute it with 117.7 mL of water to obtain a total volume of 128.4 mL of 0.50 M HCl solution.

Step-by-step solution

Calculate the needed mass of solute for (a)

We know, Molarity (M) = moles of solute/liters of solution. First, we have to find the moles of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ needed for 1.50 L of solution. Then, convert the moles to mass using the molecular weight. Moles of solute = Molarity * Volume of solution = 0.110 M * 1.50 L = 0.165 mol Now, we find the molecular weight of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$: (2 * (14.01 (N) + 4 * 1.008 (H))) + 32.07 (S) + 4 * 16 (O) ≈ 132.14 g/mol Mass of solute needed = moles of solute * molecular weight = 0.165 mol * 132.14 g/mol ≈ 21.80 g

Preparing the solution (a)

To prepare 1.50 L of a 0.110 M solution of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$, weigh 21.80 g of solid ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ and dissolve it completely in water. Then, add water until the total volume of the solution is exactly 1.50 L.

Calculate the needed mass of solute for (b)

We know, Molality (m) = moles of solute/kilograms of solvent. First, we need to find the moles of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ needed to achieve a 0.65 m solution in 225 g of solution. Then, convert the moles to mass using the molecular weight. Moles of solute = Molality * mass of solvent (kg) = 0.65 m * (225 g /1000) ≈ 0.14625 mol Now, we find the molecular weight of ${\mathrm{Na}}_2{\mathrm{CO}}_3$: (2 * 22.99 (Na)) + 12.01 (C) + 3 * 16 (O) ≈ 105.99 g/mol Mass of solute needed = moles of solute * molecular weight = 0.14625 mol * 105.99 g/mol ≈ 15.5 g

Preparing the solution (b)

For part (b), weigh 15.5 g of solid ${\mathrm{Na}}_2{\mathrm{CO}}_3$ and dissolve it completely in water. Then, add water until the total mass of the solution is exactly 225 g.

Calculate the needed mass of solute for (c)

For part (c), we have mass-volume percent and the density. The mass of the solute in a given volume can be calculated using the mass-volume percent. First, we need to find the mass of the solution: Mass of solution = Volume of solution * Density = 1.20 L * (1.16 g/mL) * (1000 mL/L) = 1392 g Now, calculate the mass of solute using mass-volume percent using the formula: Mass of solute = (mass-volume percent) * (mass of solution) Mass of solute = 15.0% * 1392 g ≈ 208.8 g

Preparing the solution (c)

For part (c), weigh 208.8 g of solid solute $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ and dissolve it completely in water. Then, add water until the total mass of the solution is 1392 g.

Determine the volume of 6 M HCl needed to neutralize 5.5 g of $\mathrm{Ba}(\mathrm{OH}{)}_2$ (d)

First, we calculate the moles of $\mathrm{Ba}(\mathrm{OH}{)}_2$ using its molecular weight: (137.33 (Ba) + 2 * (15.999 (O) + 1.008 (H))) ≈ 171.35 g/mol Moles of $\mathrm{Ba}(\mathrm{OH}{)}_2$ = mass / molecular weight = 5.5 g / 171.35 g/mol ≈ 0.0321 mol The balanced chemical equation for the neutralization reaction is: $\mathrm{Ba}(\mathrm{OH}{)}_2+2\mathrm{HCl}\to {\mathrm{BaCl}}_2+2{\mathrm{H}}_2\mathrm{O}$ From the balanced equation, we find that 1 mol of $\mathrm{Ba}(\mathrm{OH}{)}_2$ reacts with 2 moles of HCl. So, moles of HCl needed = 2 * 0.0321 mol ≈ 0.0642 mol Then, calculate the volume of 6 M HCl required to provide the needed moles of HCl: Volume (L) = moles / molarity = 0.0642 mol / 6 M ≈ 0.0107 L or 10.7 mL

Preparing the solution (d)

For part (d), take 10.7 mL of 6 M HCl and carefully dilute it with water to create a 0.50 M solution that would exactly neutralize the 5.5 g of $\mathrm{Ba}(\mathrm{OH}{)}_2$. First, calculate the total volume of the 0.50 M solution: Total Volume (L) = moles of HCl / molarity = 0.0642 mol / 0.50 M ≈ 0.1284 L or 128.4 mL This means that the water required to dilute 10.7 mL of 6 M HCl is 128.4 mL - 10.7 mL ≈ 117.7 mL. So, add 117.7 mL of water to the 10.7 mL of 6 M HCl to get the 0.50 M HCl solution.

Key concepts

Molarity and Preparation of Solutions

Understanding molarity is essential when preparing aqueous solutions in a lab environment. Molarity, represented by the symbol 'M', is a measure of concentration that describes the number of moles of a solute per liter of solution. To find the molarity of a solution, you use the formula:
$$M=\frac{\text{moles of solute}}{\text{liters of solution}}$$.
For example, if you need to prepare a certain volume of a molar solution, you would first calculate the number of moles required. After finding the molarity, you multiply the volume of the solution by the calculated molar concentration. Once you've determined the necessary moles, you can convert this to mass using the molecular weight of the solute.
To prepare the solution, the calculated mass of the solute is dissolved in a smaller volume of solvent, and then more solvent is added until the total volume reaches the desired amount. This method ensures that the solution achieves the correct concentration. Additionally, using analytical balances for measuring mass and volumetric flasks for measuring out solvents can improve the accuracy of your preparation.

Molality and Solution Concentration

Molality, denoted as 'm', is another concentration term but it measures the moles of solute per kilogram of solvent. Unlike molarity, molality is not affected by changes in temperature, as it depends only on the mass of the solvent.
The formula for calculating molality is:
$$m=\frac{\text{moles of solute}}{\text{kilograms of solvent}}$$.
When preparing a solution with a specific molality, you need to consider the mass of the solvent rather than the volume of the solution. This makes molality particularly useful in scenarios where temperature variations are significant, as volume can expand or contract with temperature. After determining the moles of the solute as per the desired molality, you proceed by adding the solute to the measured mass of the solvent before making up the final mass. It's crucial to ensure that you're using pure solvent and guarantee the accuracy of the mass measurements for precise results.

Mass-Volume Percent Composition

The mass-volume percent is a useful concentration metric, especially when dealing with solutions where both the solute and solvent are significant to the mixture's final properties. Defined as the mass of solute per 100 mL of solution, this measure is expressed as a percentage.
To calculate the mass of the solute using the mass-volume percent composition:
$$\text{Mass of solute (g)}=\left(\text{mass-volume percent}\right)\times \left(\text{mass of solution (g)}\right)$$.
Once you've calculated the mass of the solute using the formula above, it's important to slowly add the solute to the solvent to prevent any sudden reactions or splashing. Make sure to mix the solution until the solute is fully dissolved. The use of a top-loading balance for measuring out larger quantities of solutes and a graduated cylinder or a volumetric flask for measuring the solvent can be beneficial for preparing accurate mass-volume percent compositions.

Neutralization Reactions

In chemistry, neutralization reactions are reactions where an acid and a base react to form water and a salt. These reactions are important in preparing solutions where precise pH adjustments are needed. For a neutralization reaction, the equation is typically:
$$\text{Acid}+\text{Base}\to \text{Salt}+\text{Water}$$.
The amounts of acid and base required to achieve neutralization are based on their molar stoichiometry. To calculate the volume of an acid solution (such as HCl) needed to neutralize a given amount of base (like Ba(OH)₂), you must first determine the moles of base present. Then, based on the balanced chemical equation, you can calculate the necessary moles of acid. Knowing the molarity of the acid solution, you can find the volume required using the formula:
$$\text{Volume (L)}=\frac{\text{moles of acid}}{\text{molarity}}$$.
Once the required volume of acid is known, it can be carefully added to a volumetric vessel containing the base. The end-point of a neutralization can be determined using pH indicators or pH meters to confirm a neutral pH, typically around 7. In the context of preparing a specific molarity solution, like in the ORIGINAL EXERCISE AND SOLUTION section above, this understanding allows for precise dilutions and mixture preparations.