What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M NH3 with 25.00 mL of 0.10 M NH4Cl? Assume that the volume of the solutions are additive and that Kb= 1.8 × 10-5 for NH3.
The Henderson-Hasselbalch equation is
pH = pKa + log(base)/(acid)
Remember to change Kb to pKa.
pKa + pKb = pKw = 14
If you convert Kb to pKb you can use the above to convert that to pKa.
9.56
To find the pH of the solution prepared by mixing NH3 and NH4Cl, we need to consider the acid-base equilibrium between these two substances.
NH3 is a weak base, while NH4Cl is the salt formed by the reaction of NH3 with HCl, a strong acid. When NH4Cl is dissolved in water, it dissociates into NH4+ and Cl- ions. The NH4+ ions can act as an acidic species by donating a proton.
The equilibrium reaction between NH3 and NH4+ can be represented as follows:
NH3 + H2O ⇌ NH4+ + OH-
In this equation, NH4+ acts as an acid (donating a proton) and NH3 acts as a base (accepting a proton).
Since we know the concentrations of NH3 and NH4Cl, we can use the concept of moles and volumes to find the moles of NH3 and NH4+ in the solution.
Moles of NH3 = Concentration of NH3 (0.10 M) × Volume of NH3 (50.00 mL) = 0.10 mol/L × 0.0500 L = 0.00500 mol
Moles of NH4+ = Concentration of NH4Cl (0.10 M) × Volume of NH4Cl (25.00 mL) = 0.10 mol/L × 0.0250 L = 0.00250 mol
Next, we need to calculate the concentration of OH- ions in the solution. Since NH3 is a weak base, it reacts with water to form OH- ions. The reaction is as follows:
NH3 + H2O ⇌ NH4+ + OH-
The concentration of OH- ions can be determined using the Kb value of NH3, which represents the equilibrium constant for the reaction of NH3 with water. In this case, Kb = 1.8 × 10^(-5).
Kb = [NH4+][OH-] / [NH3]
Given the moles of NH3 and NH4+ calculated earlier, we can rearrange this equation to solve for [OH-]:
[OH-] = (Kb × [NH3]) / [NH4+]
[OH-] = (1.8 × 10^(-5) × 0.00500 mol) / 0.00250 mol
[OH-] = 0.036 mol/L
Now, we can find the pOH of the solution by taking the negative logarithm of the OH- concentration:
pOH = -log [OH-] = -log (0.036) = 1.44
Finally, we can find the pH of the solution using the relationship:
pH + pOH = 14
pH = 14 - pOH = 14 - 1.44 = 12.56
Therefore, the pH of the solution prepared by mixing 50.00 mL of 0.10 M NH3 with 25.00 mL of 0.10 M NH4Cl is approximately 12.56.