Q.Calculate the pH of a 0.10 M ammonia solution. Calculate the pH after 50.0 mL of this solution is treated with 25.0 mL of 0.10 M HCl. The dissociation constant of ammonia, Kb = 1.77 × 10⁻⁵.

Concept: Weak base equilibrium, then buffer solution pH after partial neutralization.

Part 1: Initial pH of 0.10 M ammonia

Ammonia establishes equilibrium: $\ce{NH3 + H2O <=> NH4+ + OH-}$ with $K_b = 1.77 \times 10^{-5}$.

Setting up the ICE table with $x = [\ce{OH-}]$:

So $\text{pOH} = -\log(1.33 \times 10^{-3}) = 2.88$, giving $\text{pH} = 14 - 2.88 = 11.12$.

Part 2: pH after adding HCl

Moles: $\ce{NH3} = 0.10 \times 0.050 = 5.0 \times 10^{-3}$ mol; $\ce{HCl} = 0.10 \times 0.025 = 2.5 \times 10^{-3}$ mol.

The neutralization $\ce{NH3 + HCl -> NH4Cl}$ consumes 2.5 mmol of each, leaving:

• $\ce{NH3}$: $5.0 - 2.5 = 2.5$ mmol
• $\ce{NH4+}$: $2.5$ mmol (formed) …

Ammonia in water establishes an equilibrium as a weak base:

The base dissociation constant $K_b = 1.77 \times 10^{-5}$ tells us how far this equilibrium lies to the right. When we add a strong acid like HCl, it donates protons that convert ammonia into ammonium ion, and if we don't add enough acid to consume all the ammonia, we end up with a mixture of $\text{NH}_3$ and $\text{NH}_4^+$ — a buffer solution that resists pH change. The key is to track moles before and after the reaction, then apply the appropriate equilibrium expression.

Part 1: pH of 0.10 M ammonia solution

1. Set up the equilibrium table.

   Let $x$ be the concentration of $\text{OH}^-$ produced at equilibrium.

   Species	Initial (M)	Change (M)	Equilibrium (M)
   $\text{NH}_3$	0.10	$-x$	$0.10 - x$
   $\text{NH}_4^+$	0	$+x$	$x$
   $\text{OH}^-$	0	$+x$	$x$

2. Write the $K_b$ expression and assume $x \ll 0.10$.

Solving for $x$:

Check: $\frac{1.33 \times 10^{-3}}{0.10} \times 100\% = 1.33\%$, well under 5%, so the approximation holds.

3. Convert $[\text{OH}^-]$ to pH.

Part 2: pH after adding 25.0 mL of 0.10 M HCl to 50.0 mL of 0.10 M $\text{NH}_3$

4. Calculate initial moles of each species.

   • Moles of $\text{NH}_3 = 0.10 \, \text{M} \times 0.0500 \, \text{L} = 5.0 \times 10^{-3} \, \text{mol}$
   • Moles of HCl $= 0.10 \, \text{M} \times 0.0250 \, \text{L} = 2.5 \times 10^{-3} \, \text{mol}$

5. The neutralization reaction goes to completion.

We now have equal moles of the weak base $\text{NH}_3$ and its conjugate acid $\text{NH}_4^+$ — a buffer at the half-equivalence point.

6. Find the total volume and concentrations. …