Ammonia (NH3) ionizes according to the following reaction:
NH3(aq)^+ H2O(l) ⇌ NH4^+(aq) + OH^–(aq)
The base dissociation constant for ammonia (NH3) is Kb = 1.8 × 10^–5. Ammonia (NH3) also has a chloride salt, ammonium chloride (NH4Cl), which is soluble in water.
If 0.070 M of ammonia (NH3) and 0.035 M of its salt ammonium chloride (NH4Cl) are mixed in a solution, what is the pH of this solution?
A weak base and its salt is a buffer. Use the Henderson-Hasselbalch equation.
pH = pKa + log (base)/(acid)
Don't have pKa. You have Kb, convert to pKb, then pKa + pKb = pKw = 14 which will let you solve for pKa.
To find the pH of the solution, we need to determine the concentrations of both OH^- and H3O^+ ions in the solution. Since ammonia is a weak base, it will generate OH^- ions upon ionization.
First, let's calculate the concentration of OH^- ions using the base dissociation constant (Kb) for ammonia:
Kb = [NH4^+][OH^-] / [NH3]
From the reaction equation, we know that [NH4^+] = [OH^-], so we can substitute this into the equation:
Kb = [OH^-]^2 / [NH3]
We can rearrange the equation to solve for [OH^-]:
[OH^-] = √(Kb × [NH3])
Now, plug in the given values into the equation:
Kb = 1.8 × 10^–5
[NH3] = 0.070 M
[OH^-] = √((1.8 × 10^–5) × (0.070))
Next, calculate the concentration of NH4+ ions from the concentration of ammonium chloride:
[NH4^+] = 0.035 M
The concentration of H3O^+ ions can be obtained since H3O^+ and OH^- are related by the equilibrium constant (Kw) for water:
Kw = [H3O^+][OH^-] = 1.0 × 10^-14
Since the concentration of OH^- can be determined, we can solve for [H3O^+]:
[H3O^+] = Kw / [OH^-]
Substitute the given value for Kw and the calculated value for [OH^-]:
[H3O^+] = (1.0 × 10^-14) / [OH^-]
Finally, to find the pH of the solution, use the relationship between pH and [H3O^+]:
pH = -log[H3O^+]
Plug in the calculated value for [H3O^+] to find the pH.