Question

Calculate the volume of \(0.150-\mathrm{M} \mathrm{HCl}\) required to titrate to the equivalence point for each of these samples. (a) \(25.0 \mathrm{~mL}\) of \(0.175-\mathrm{M} \mathrm{KOH}\) (b) \(15.0 \mathrm{~mL}\) of \(6.00-\mathrm{M} \mathrm{NH}_{3}\) (c) \(15.0 \mathrm{~mL}\) of propylamine, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\), which has a density of \(0.712 \mathrm{~g} / \mathrm{mL}\) (d) \(40.0 \mathrm{~mL}\) of \(0.0050-\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\)

Short answer

(a) 29.2 mL, (b) 600 mL, (c) 1203 mL, (d) 2.67 mL.

Step-by-step solution

Determine moles of KOH for (a)

Calculate moles of KOH using the provided molarity and volume:\[ \text{moles of KOH} = M \times V = 0.175 \, \text{mol/L} \times 0.0250 \, \text{L} = 0.004375 \, \text{mol} \]

Calculate volume of HCl for (a)

Use the balanced equation of reaction: HCl + KOH → KCl + H2O. Since the ratio is 1:1, moles of HCl needed are the same as moles of KOH:\[ V(\text{HCl}) = \frac{\text{moles of HCl}}{\text{Molarity of HCl}} = \frac{0.004375 \, \text{mol}}{0.150 \, \text{mol/L}} = 0.0292 \, \text{L} \text{ (or 29.2 mL)} \]

Determine moles of NH3 for (b)

Calculate moles of NH3 using its molarity and volume:\[ \text{moles of NH}_{3} = M \times V = 6.00 \, \text{mol/L} \times 0.0150 \, \text{L} = 0.090 \, \text{mol} \]

Calculate volume of HCl for (b)

Use the balanced equation NH3 + HCl → NH4Cl. The stoichiometry is 1:1, so moles of HCl = moles of NH3:\[ V(\text{HCl}) = \frac{\text{moles of HCl}}{\text{Molarity of HCl}} = \frac{0.090 \, \text{mol}}{0.150 \, \text{mol/L}} = 0.600 \, \text{L} \text{ (or 600 mL)} \]

Calculate moles of propylamine for (c)

Determine the mass and then moles of propylamine:Mass = Volume × Density = 15.0 mL × 0.712 g/mL = 10.68 g\[ \text{Molar mass of CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{NH}_{2} = 3(12.01) + 8(1.01) + 14.01 = 59.11 \, \text{g/mol} \]\[ \text{moles of CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{NH}_{2} = \frac{10.68 \, \text{g}}{59.11 \, \text{g/mol}} = 0.1805 \, \text{mol} \]

Calculate volume of HCl for (c)

Using the 1:1 stoichiometry:\[ V(\text{HCl}) = \frac{0.1805 \, \text{mol}}{0.150 \, \text{mol/L}} = 1.203 \, \text{L} \text{ (or 1203 mL)} \]

Determine moles of Ba(OH)2 for (d)

Calculate moles of Ba(OH)2 and account for dissociation:\[ \text{moles of Ba(OH)}_2 = M \times V = 0.0050 \, \text{mol/L} \times 0.0400 \, \text{L} = 0.0002 \, \text{mol} \]Each mole of Ba(OH)2 provides 2 moles of OH⁻ ions, so\[ \text{equivalent moles of OH⁻} = 2 \times 0.0002 \, \text{mol} = 0.0004 \, \text{mol} \]

Calculate volume of HCl for (d)

Use the reaction 2HCl + Ba(OH)2 → BaCl2 + 2H2O to find volume:\[ V(\text{HCl}) = \frac{0.0004 \, \text{mol}}{0.150 \, \text{mol/L}} = 0.0027 \, \text{L} \text{ (or 2.67 mL)} \]

Key concepts

Molarity

Molarity is a fundamental concept in chemistry that measures the concentration of a solution. It is defined as the number of moles of a solute divided by the volume of the solution in liters. The formula is typically expressed as:\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]When performing titration calculations, knowing the molarity of the solutions involved is crucial because it allows for the determination of the amount of reactant required to reach a specific point in the chemical reaction, such as the equivalence point. For instance, in the case of titrating 0.150-M HCl with substances like KOH, NH3, propylamine, or Ba(OH)2, the molarity of HCl helps calculate the exact volume needed to neutralize the base.

Equivalence point

The equivalence point in a titration is where the amount of titrant added is just enough to completely neutralize the analyte solution. In other words, it is the point at which the number of moles of titrant equals the number of moles of the substance being titrated, based on their molar ratio from the balanced chemical equation. The significance of reaching the equivalence point during a titration is that it accurately allows us to determine the concentration of an unknown solution when the concentration of the other is known. In titration exercises, identifying the equivalence point ensures that the calculations are precise and that the chemical equation's stoichiometry is respected. For example, in the titration of HCl with KOH, the equivalence point is when one mole of KOH has reacted with one mole of HCl to form water and KCl, following their 1:1 stoichiometric ratio.

Chemical reactions

Chemical reactions involve the transformation of reactants into products. They are represented by balanced chemical equations that illustrate the ratios in which substances react and are produced. During a titration process, it is crucial to rely on these balanced reactions to determine how much of each reactant is required.In a typical titration problem, such as one involving HCl and KOH, the balanced chemical equation is:\[ \text{HCl (aq) + KOH (aq) } \rightarrow \text{ KCl (aq) + H}_2\text{O (l)} \]This equation shows a 1:1 reaction ratio, meaning one mole of hydrochloric acid reacts with one mole of potassium hydroxide. Understanding the concept of balanced chemical reactions helps solve titration problems by utilizing the correct molar ratios to compute the necessary volumes or concentrations.

Stoichiometry

Stoichiometry is an essential concept in chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. It uses the coefficients in a balanced chemical equation to establish the ratios of moles of reactants and products.During a titration, stoichiometry helps determine the exact amounts of titrant needed to reach the equivalence point. Using stoichiometry, one can calculate the moles of an unknown solution given the volume and concentration of a titrant. For example, when titrating HCl with Ba(OH)2, the stoichiometrically balanced equation is:\[ \text{2HCl (aq) + Ba(OH)}_2\text{ (aq)} \rightarrow \text{BaCl}_2\text{ (aq) + 2H}_2\text{O (l)} \]Here, two moles of HCl are required for each mole of Ba(OH)2, illustrating the importance of stoichiometric ratios in calculating the correct volume of HCl needed.