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A 20.00 ml sample of 0.150 M NH₃ is being titrated with 0.200 M HCl. What is the pH after 5.00 ml of HCl has been added? Kb of NH₃ = 1.8 × 10⁻⁵.

User BeardFist
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Final answer:

To determine the pH after adding 5.00 mL of 0.200 M HCl to a 20.00 mL sample of 0.150 M NH₃, calculate the moles of NH₃ and HCl initially present. Then, determine the resulting concentration of NH₄⁺ and calculate the pH using the expression -log[H₃O⁺].

Step-by-step explanation:

To determine the pH after adding 5.00 mL of 0.200 M HCl to a 20.00 mL sample of 0.150 M NH₃, we need to calculate the moles of NH₃ and HCl initially present, and then determine the resulting concentration of NH₄⁺. Finally, we can calculate the pH using the expression -log[H₃O⁺].

First, calculate the moles of NH₃: Moles of NH₃ = Molarity × Volume = 0.150 M × 20.00 mL = 0.0030 moles.

Next, calculate the moles of HCl: Moles of HCl = Molarity × Volume = 0.200 M × 5.00 mL = 0.0010 moles.

The reaction between NH₃ and HCl is as follows: NH₃ + HCl → NH₄Cl. Since HCl is a strong acid, it completely ionizes to produce H₃O⁺. Therefore, the moles of NH₄⁺ formed will be equal to the moles of HCl added: 0.0010 moles.

The resulting concentration of NH₄⁺ is given by the expression: [NH₄⁺] = moles ÷ total volume = 0.0010 moles ÷ (20.00 mL + 5.00 mL) = 0.025 M.

Finally, calculate the pH using the expression: pH = -log[H₃O⁺] = - log(0.025 M) = 1.60.

User Kiddouk
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