What is the pH of a solution prepared by mixing 10 mL of 0.10 M NH3 and 5.0 mL of 0.24 M NH4+. Assume Ka(NH4+) = 5.56 x 10^-10
Use the Henderson-Hasselbalch equation.
To find the pH of the solution, we need to determine the concentration of the hydronium ion (H3O+).
First, let's calculate the moles of NH3 and NH4+ in each solution:
Moles of NH3 = volume (L) * concentration (M)
= 0.010 L * 0.10 M
= 0.001 mol
Moles of NH4+ = volume (L) * concentration (M)
= 0.005 L * 0.24 M
= 0.0012 mol
Since NH3 is a weak base and NH4+ is its conjugate acid, we can consider the acid-base reaction between them:
NH3 + H2O ⇌ NH4+ + OH-
From the reaction, we can see that NH4+ reacts with water to produce OH-. This means that the concentration of OH- will change.
To determine the concentration of OH-, we need to calculate the moles of OH- added to the solution. Since NH3 and NH4+ have a 1:1 stoichiometry, the number of moles of OH- formed will be equal to the number of moles of NH4+:
Moles of OH- = 0.0012 mol
Since OH- is a strong base, it will react with water to produce H2O and OH-. This reaction will consume OH-, resulting in a decrease in its concentration.
H2O + OH- ⇌ H3O+ + OH-
From this reaction, we can see that OH- reacts with water to produce H3O+. This means that the concentration of H3O+ will change.
To determine the concentration of H3O+, we need to calculate the moles of H3O+ formed. Since OH- and H3O+ have a 1:1 stoichiometry, the number of moles of H3O+ formed will also be equal to the number of moles of OH-:
Moles of H3O+ = 0.0012 mol
To find the concentration of H3O+, we divide the moles of H3O+ by the total volume of the solution:
Total volume of the solution = volume of NH3 + volume of NH4+
= 0.010 L + 0.005 L
= 0.015 L
Concentration of H3O+ = Moles of H3O+ / Total volume of the solution
= 0.0012 mol / 0.015 L
= 0.08 M
Now we have the concentration of the hydronium ion (H3O+), which indicates the acidity of the solution. Since we're looking for the pH, we can use the formula:
pH = -log[H3O+]
Substituting the value we calculated:
pH = -log(0.08)
= 1.10
Therefore, the pH of the solution prepared by mixing 10 mL of 0.10 M NH3 and 5.0 mL of 0.24 M NH4+ is approximately 1.10.