Question

How many milliliters of 0.105 \(\mathrm{M}\) HCl are needed to titrate each of the following solutions to the equivalence point: (a) 45.0 \(\mathrm{mL}\) of \(0.0950 \mathrm{MNaOH},(\mathbf{b}) 22.5 \mathrm{mL}\) of \(0.118 \mathrm{MNH}_{3},(\mathbf{c}) 125.0\) mL of a solution that contains 1.35 gof NaOH perliter?

Short answer

(a) 40.7 mL of 0.105 M HCl is needed to titrate 45.0 mL of 0.0950 M NaOH to the equivalence point. (b) 25.3 mL of 0.105 M HCl is needed to titrate 22.5 mL of 0.118 M NH3 to the equivalence point. (c) 40.18 mL of 0.105 M HCl is needed to titrate 125.0 mL of a solution containing 1.35 g of NaOH per liter to the equivalence point.

Step-by-step solution

Step 1:Find moles of NaOH

To find the moles of NaOH in the solution, use the formula moles of solute = molarity × volume. First, convert the volume to liters: 45.0 mL = 0.0450 L. Then, calculate the moles of NaOH: moles of NaOH = 0.0950 M × 0.0450 L = 0.004275 moles

Step 2:Find volume of 0.105 M HCl needed

Now, use the moles of NaOH and the molarity of HCl to find the volume of HCl required: moles of HCl = moles of NaOH 0.004275 moles = 0.105 M × volume of HCl in L volume of HCl in L = 0.040714 L Finally, convert the volume of HCl to milliliters: 0.040714 L × 1000 = 40.7 mL. Therefore, 40.7 mL of 0.105 M HCl is needed to titrate 45.0 mL of 0.0950 M NaOH to the equivalence point. (b) 22.5 mL of 0.118 M NH3

Step 1:Find moles of NH3

First, convert the volume to liters: 22.5 mL = 0.0225 L. Then, calculate the moles of NH3: moles of NH3 = 0.118 M × 0.0225 L = 0.002655 moles

Step 2:Find volume of 0.105 M HCl needed

Use the moles of NH3 and the molarity of HCl to find the volume of HCl required: moles of HCl = moles of NH3 0.002655 moles = 0.105 M × volume of HCl in L volume of HCl in L = 0.025286 L Finally, convert the volume of HCl to milliliters: 0.025286 L × 1000 = 25.3 mL. Therefore, 25.3 mL of 0.105 M HCl is needed to titrate 22.5 mL of 0.118 M NH3 to the equivalence point. (c) 125.0 mL of a solution that contains 1.35 g of NaOH per liter

Step 1:Find moles of NaOH

First, find the moles of NaOH contained in 1.00 L of the solution: moles of NaOH in 1.00 L = (1.35 g NaOH) / (40.00 g/mol)= 0.03375 moles/L (since the molar mass of NaOH is 40.00 g/mol) Now, find the moles of NaOH in 0.125 L of the solution since 125.0 mL = 0.125 L: moles of NaOH = 0.03375 moles/L × 0.125 L = 0.00421875 moles

Step 2:Find volume of 0.105 M HCl needed

Use the moles of NaOH and the molarity of HCl to find the volume of HCl required: moles of HCl = moles of NaOH 0.00421875 moles = 0.105 M × volume of HCl in L volume of HCl in L = 0.0401786 L Finally, convert the volume of HCl to milliliters: 0.0401786 L × 1000 = 40.18 mL (rounded to two decimal places). Therefore, 40.18 mL of 0.105 M HCl is needed to titrate 125.0 mL of a solution containing 1.35 g of NaOH per liter to the equivalence point.

Key concepts

Titration Equivalence Point

Understanding the titration equivalence point is crucial in a titration process. A titration is an analytical procedure in which a solution of known concentration, called the titrant, is gradually added to a volume of another solution of unknown concentration until the chemical reaction between the two solutions is completed. The point at which the added titrant is stoichiometrically equivalent to the number of moles of substance in the sample is the equivalence point. In a typical acid-base titration, the equivalence point is achieved when the number of moles of hydrogen ions (H+) equals the number of moles of hydroxide ions (OH-).

This is a pivotal moment in titration as it determines when the reaction has been completely neutralized. To interpret the equivalence point accurately, indicators or pH meters are often used. The indicator chosen must have a transition range that includes the pH at which the equivalence point occurs. For acid-base titrations, the equivalence point typically occurs around a pH of 7. However, the exact pH can vary depending on the strength of the acids and bases involved.

Improvements in understanding the equivalence point can be made by performing the titration slowly, and with careful monitoring, to observe the color change or pH shift that indicates the equivalence point has been reached.

Molarity Concentration

The concept of molarity concentration is essential for solving many chemistry problems, including titrations. Molarity, denoted by M, is a measure of concentration in chemistry that describes the amount of solute (substance being dissolved) contained in a given volume of solution. It is expressed as the number of moles of solute per liter of solution (mol/L).

To calculate molarity, you can use the formula: \[ Molarity = \frac{moles \, of \, solute}{volume \, of \, solution \, in \, liters} \]
When working with solutions, it's often required to convert volumes to liters to align with the units used in molarity. In titration problems, knowing the molarity of each reactant allows us to determine how much of one solution is required to react completely with another. The exercise provided demonstrates the use of molarity to find out how many milliliters of a hydrochloric acid solution are needed to titrate various bases to their equivalence points, showcasing the importance of this concentration measure in practical scenarios.

Stoichiometry

Stoichiometry lies at the heart of the chemical equations used in titration calculations. It refers to the calculation of reactants and products in chemical reactions and is based on the law of conservation of mass. Stoichiometry allows us to predict the amounts of substances consumed and produced in a given reaction.

When applying stoichiometry to titration calculations, the mole ratio of the reactants, derived from the balanced chemical equation, becomes invaluable. For instance, in an acid-base titration, the stoichiometry often involves a 1:1 mole ratio of acid to base. Using the known molarity and volume of one reactant, you can calculate the moles of that reactant, and then use the stoichiometry to find the required volume of the other reactant.

To improve comprehension in this area, focusing on balancing chemical equations and understanding mole relationships is critical. Students should practice setting up conversion factors based on the coefficients in balanced equations, as this is the foundation for stoichiometry calculations in titration and many other chemical processes.