A 20.00 mL sample of 0.150 M NH3 is being titrated with 0.200 M HCl. What is the pH after 10.00 mL of HCl has been added? Kb of NH3 = 1.8 × 10−5

1. Break Down the Problem

To find the pH after adding 10.00 mL of 0.200 M HCl to 20.00 mL of 0.150 M NH₃, we will:

1. Calculate the moles of NH₃ and HCl.
2. Determine the moles of NH₃ remaining after the reaction with HCl.
3. Calculate the concentration of the resulting NH₄⁺ (from NH₃ reacting with HCl).
4. Use the Kb and the concentration of NH₄⁺ to find the resulting pH.

2. Relevant Concepts

1. Moles of a solution is calculated using: $\text{Moles} = \text{Concentration (M)} \times \text{Volume (L)}$
2. The reaction between NH₃ and HCl is: $\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4^+ + \text{Cl}^-$
3. The relation between Kb and Ka: $K_w = K_a \times K_b$ Where $K_w = 1.0 \times 10^{-14}$.
4. pH calculation involves finding the concentration of hydrogen ions.

3. Analysis and Detail

1. Calculate Moles of NH₃: $\text{Moles of NH}_3 = 0.150 \, \text{M} \times 0.0200 \, \text{L} = 0.00300 \, \text{moles}$

2. Calculate Moles of HCl: $\text{Moles of HCl} = 0.200 \, \text{M} \times 0.0100 \, \text{L} = 0.00200 \, \text{moles}$

3. Determine Moles after Reaction:

   • Moles of NH₃ remaining: $\text{Remaining NH}_3 = 0.00300 \, \text{moles} - 0.00200 \, \text{moles} = 0.00100 \, \text{moles}$
   • Moles of NH₄⁺ produced is equal to the moles of HCl: $\text{Moles of NH}_4^+ = 0.00200 \, \text{moles}$

4. Calculate the Total Volume: $\text{Total Volume} = 20.00 \, \text{mL} + 10.00 \, \text{mL} = 30.00 \, \text{mL} = 0.0300 \, \text{L}$

5. Concentration of NH₃ and NH₄⁺:

   • Concentration of NH₃: $[\text{NH}_3] = \frac{0.00100 \, \text{moles}}{0.0300 \, \text{L}} = 0.0333 \, \text{M}$
   • Concentration of NH₄⁺: $[\text{NH}_4^+] = \frac{0.00200 \, \text{moles}}{0.0300 \, \text{L}} = 0.0667 \, \text{M}$

6. Calculate Ka for NH₄⁺: $K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10}$

7. Use Ka to find [H⁺]: For the dissociation of NH₄⁺: $K_a = \frac{[\text{H}^+][\text{NH}_3]}{[\text{NH}_4^+]} \implies 5.56 \times 10^{-10} = \frac{x (0.0333)}{0.0667}$

   • Cross-multiplying gives: $5.56 \times 10^{-10} \times 0.0667 = 0.0333x \implies x = \frac{3.71 \times 10^{-11}}{0.0333} \approx 1.11 \times 10^{-9}$
   • This $x$ is [H⁺].

8. Finding pH: $pH = -\log(1.11 \times 10^{-9}) \approx 8.95$

4. Verify and Summarize

• The calculations show that the moles of NH₃ and HCl were accurately derived and the concentrations calculated.
• The $K_a$ relationship and resulting $H^+$ concentration leading to the pH were verified.

Final Answer

The pH after adding 10.00 mL of 0.200 M HCl is approximately 8.95.