Answer:
Explanation:
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
