What is the pH of the solution obtained by diluting 500 mL of 0.0200 M NH3 (aq) to 1.000 L ?
NH3 + HOH ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3)
This problem is like the CH3NH2 problem I worked for you. The only difference is that it is a diluted solution. The new concn of NH3 is 0.0200 M x (500/1000) = ??
To find the pH of the diluted solution, we need to determine the concentration of NH3 after dilution.
Step 1: Calculate the moles of NH3 in the original solution.
The molarity (M) of the original solution is given as 0.0200 M.
To calculate the moles of NH3, we can use the formula:
moles = molarity × volume
moles = 0.0200 M × 0.500 L
moles = 0.0100 mol
Step 2: Calculate the new concentration of NH3 after dilution.
After dilution, the total volume is 1.000 L, and the original volume was 0.500 L. Therefore, the dilution factor is:
dilution factor = final volume / initial volume
dilution factor = 1.000 L / 0.500 L
dilution factor = 2
Since the dilution factor is 2, the concentration of NH3 in the diluted solution is:
new concentration = original concentration / dilution factor
new concentration = 0.0200 M / 2
new concentration = 0.0100 M
Step 3: Calculate the pOH of the solution.
Since NH3 is a weak base, we can use the relationship between pOH and concentration of OH- ions:
pOH = -log[OH-]
NH3 reacts with water to produce OH- ions:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Since the stoichiometry is 1:1, the concentration of OH- ions is equal to the concentration of NH3 in the diluted solution:
[OH-] = 0.0100 M
Using the concentration of OH-, we can calculate the pOH:
pOH = -log(0.0100)
pOH = 2
Step 4: Calculate the pH of the solution.
The pH and pOH are related through the equation:
pH + pOH = 14
Therefore, the pH of the solution is:
pH = 14 - pOH
pH = 14 - 2
pH = 12
So, the pH of the solution obtained by diluting 500 mL of 0.0200 M NH3 (aq) to 1.000 L is 12.