Final answer:
The pH of the solution formed when hydrochloric acid is added to NH₃ is approximately 1.659.
Step-by-step explanation:
To calculate the pH of the solution formed when 45.0 mL of hydrochloric acid (HCl) is added to 50.0 mL of NH₃, we need to consider the reaction between HCl and NH₃. HCl is a strong acid that completely dissociates in water, while NH₃ is a weak base that partially ionizes in water. The reaction can be represented as follows:
HCl + NH₃ → NH₄Cl
First, we need to calculate the moles of HCl and NH₃. Moles = Molarity x Volume (in L).
Moles of HCl = 0.213 M x 0.045 L = 0.009585 mol
Moles of NH₃ = 0.150 M x 0.050 L = 0.0075 mol
Next, we determine the excess reactant. In this case, HCl is in excess because it has more moles.
The excess moles of HCl = Moles of HCl - Moles of NH₃ = 0.009585 mol - 0.0075 mol = 0.002085 mol
Now, we calculate the moles of the resulting salt, NH₄Cl.
Moles of NH₄Cl = 0.002085 mol
Since NH₄Cl is a strong electrolyte, it completely dissociates in water, producing NH₄⁺ and Cl⁻ ions.
Finally, we can calculate the final concentration of NH₄⁺ ions in the solution. Final concentration = Moles of NH₄⁺ / Total volume (in L).
Total volume = 45.0 mL + 50.0 mL = 95.0 mL = 0.095 L
Final concentration of NH₄⁺ = 0.002085 mol / 0.095 L = 0.021947 M
Since NH₄⁺ is the conjugate acid of NH₃, it acts as an acid in water. To calculate the pH, we use the equation pH = -log[H₃O⁺].
pH = -log(0.021947) ≈ 1.659
Therefore, the pH of the solution is approximately 1.659.