Ask Question
13 December, 16:14

Calculate the concentration of buffer components present in 295.00 mL of a buffer solution that contains 0.390 M NH4Cl and 0.390 M NH3 immediately after the addition of 1.40 mL of 6.00 M HNO3.

+3
Answers (1)
  1. 13 December, 20:09
    0
    The acid HNO₃ will react with the base NH₃ in the buffer to produce NH₄⁺, changing the concentrations of both NH₃ and NH₄⁺ in the buffer.

    Therefore, this question is solved by performing stoichiometric calculations based on those reactions:

    HNO₃ + NH₃ ⇒ NH₄⁺

    Vol HNO₃ = 1.40 mL x 1L / 1000 mL = 1.4 x 10⁻³ L

    # mol HNO₃ = 1.4 x 10⁻³ L x 6.00 mol/L = 8.4 x 10⁻³ mol HNO₃

    Since the reaction occurs in 1:1 ratio, 8.4 x 10⁻³ mol NH₃ will be consumed and 8.4 x 10⁻³ mol of NH₄⁺ will be produced.

    Initially we had

    # mol NH₄⁺ = 0.295 L x 0.390 mol/L = 0.115 mol

    # mol NH₃ = 0.115

    After reaction we will have

    # mol NH₄⁺ = 0.115 mol + 8.4 x 10⁻³ mol = 0.123 mol

    # mol NH₃ = 0.115 mol - 8.4 x 10⁻³ mol = 0.107 mol

    Therefore, their new concentrations will be

    Volume after reaction = 0.295 L + 0.0014 L = 0.296 L

    [NH₄⁺] = 0.123 mol / 0.296 L = 0.416 M

    [NH₃] = 0.107 mol / 0.296 L = 0.361 M
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Calculate the concentration of buffer components present in 295.00 mL of a buffer solution that contains 0.390 M NH4Cl and 0.390 M NH3 ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers