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.

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

the answer is d : ) have a good da

pie chart because it is in percents ( % )