Question

A 20.00 ml sample of 0.150 M NH₃ is being titrated with 0.200 M HCl. What is the pH after 25.00 ml of HCl has been added? Kb of NH₃ = 1.8 × 10⁻⁵

Answer 1

To calculate the pH after 25.00 ml of 0.200 M HCl has been added to a 20.00 ml sample of 0.150 M NH₃, we need to consider the reaction between NH₃ and HCl.

Step-by-step explanation:

To calculate the pH after 25.00 ml of 0.200 M HCl has been added to a 20.00 ml sample of 0.150 M NH₃, we need to consider the reaction between NH₃ and HCl. NH₃ is a weak base and HCl is a strong acid. The reaction between them is:

NH₃ + HCl → NH₄Cl

Since NH₃ is a weak base, it reacts with water to produce OH⁻ ions and NH₄⁺ ions. So, we can write the reaction as:

NH₄⁺ + H₂O → NH₃ + H₃O⁺

Using the concentration of NH₃ and Kb of NH₃, we can calculate the concentration of OH⁻ ions and then the pH of the solution.

1. Calculate the moles of NH₃ using the formula: moles = concentration * volume
2. Calculate the moles of OH⁻ ions produced in the reaction using the stoichiometry of the reaction.
3. Calculate the concentration of OH⁻ ions using the formula: concentration = moles / volume
4. Calculate the pOH using the formula: pOH = -log10(concentration of OH⁻)
5. Calculate the pH using the formula: pH = 14 - pOH

After following these steps, you should be able to calculate the pH after 25.00 ml of HCl has been added.

Answer 2

To calculate the pH after adding 25.00 ml of 0.200 M HCl to a 20.00 ml sample of 0.150 M NH₃, determine the moles of reactants, their reaction, the excess reagent, and then calculate the concentration of H+ ions in the solution to find the pH.

Step-by-step explanation:

A 20.00 ml sample of 0.150 M NH₃ is being titrated with 0.200 M HCl. To calculate the pH after adding 25.00 ml of HCl, we first determine the moles of NH₃ and HCl. The number of moles of NH₃ is (20.00 ml) × (0.150 M) and the number of moles of HCl is (25.00 ml) × (0.200 M). Since HCl is a strong acid, it will react completely with NH₃ in a 1:1 molar ratio to form NH₄+ and Cl-. Thus, after the reaction, we calculate the remaining moles of NH₃ (if any) or the moles of NH₄+ formed (if HCl is in excess).

For this particular scenario, after doing the calculations, all the NH₃ will be converted to NH₄+, and we will have an excess of HCl. Then, knowing the total volume of the solution and the moles of excess HCl, we calculate the concentration of H+ ions in the solution, which determines the pH of the resulting solution. The final pH will be influenced by the concentration of the additional H+ ions present from the excess HCl.