Answer :
To determine the pH of the solution after titrating 45.0 mL of 0.30 M HCl with 125.0 mL of 0.15 M aniline, follow these steps:
### Step 1: Calculate moles of HCl and aniline
Given:
- Volume of HCl ([tex]\(V_{\text{HCl}}\)[/tex]) = 45.0 mL
- Concentration of HCl ([tex]\(C_{\text{HCl}}\)[/tex]) = 0.30 M
- Volume of aniline ([tex]\(V_{\text{aniline}}\)[/tex]) = 125.0 mL
- Concentration of aniline ([tex]\(C_{\text{aniline}}\)[/tex]) = 0.15 M
Convert the volumes from mL to liters (since concentration is in M, moles/L):
[tex]\[ V_{\text{HCl}} = 45.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.045 \, \text{L} \][/tex]
[tex]\[ V_{\text{aniline}} = 125.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.125 \, \text{L} \][/tex]
Calculate moles of HCl:
[tex]\[ \text{moles HCl} = C_{\text{HCl}} \times V_{\text{HCl}} = 0.30 \, \text{M} \times 0.045 \, \text{L} = 0.0135 \, \text{moles} \][/tex]
Calculate moles of aniline:
[tex]\[ \text{moles aniline} = C_{\text{aniline}} \times V_{\text{aniline}} = 0.15 \, \text{M} \times 0.125 \, \text{L} = 0.01875 \, \text{moles} \][/tex]
### Step 2: Determine the limiting reagent and remaining moles
Reaction between HCl and aniline:
[tex]\[ \text{HCl} + \text{C}_6\text{H}_5\text{NH}_2 \rightarrow \text{C}_6\text{H}_5\text{NH}_3^+ + \text{Cl}^- \][/tex]
- Moles of HCl = 0.0135
- Moles of aniline = 0.01875
Since 1 mole of HCl reacts with 1 mole of aniline, the HCl will be the limiting reagent. Calculate remaining moles of aniline:
[tex]\[ \text{moles aniline remaining} = \text{moles aniline} - \text{moles HCl} = 0.01875 - 0.0135 = 0.00525 \, \text{moles} \][/tex]
### Step 3: Calculate the final concentrations
Total volume of the solution:
[tex]\[ V_{\text{total}} = V_{\text{HCl}} + V_{\text{aniline}} = 0.045 \, \text{L} + 0.125 \, \text{L} = 0.170 \, \text{L} \][/tex]
Final concentration of aniline:
[tex]\[ C_{\text{aniline remaining}} = \frac{\text{moles aniline remaining}}{V_{\text{total}}} = \frac{0.00525 \, \text{moles}}{0.170 \, \text{L}} = 0.0309 \, \text{M} \][/tex]
### Step 4: Calculate pH
The remaining aniline ([tex]\(C_6H_5NH_2\)[/tex]) is a weak base. Use the [tex]\(K_b\)[/tex] of aniline to determine the concentration of OH⁻ ions.
Given:
[tex]\[ K_b = 4.3 \times 10^{-10} \][/tex]
For the base dissociation of aniline:
[tex]\[ C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^- \][/tex]
Using the ICE (Initial, Change, Equilibrium) table and simplifying (since [tex]\(x \ll 0.0309\)[/tex]):
[tex]\[ K_b = \frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]} \approx \frac{x^2}{0.0309} \][/tex]
[tex]\[ x^2 = K_b \times 0.0309 \][/tex]
[tex]\[ x^2 = 4.3 \times 10^{-10} \times 0.0309 \][/tex]
[tex]\[ x^2 = 1.3287 \times 10^{-11} \][/tex]
[tex]\[ x \approx \sqrt{1.3287 \times 10^{-11}} \approx 1.152 \times 10^{-6} \][/tex]
Thus,
[tex]\[ [OH^-] = 1.152 \times 10^{-6} \, \text{M} \][/tex]
Calculate pOH:
[tex]\[ pOH = -\log [OH^-] = -\log(1.152 \times 10^{-6}) \approx 5.94 \][/tex]
Finally, calculate pH:
[tex]\[ \text{pH} = 14 - \text{pOH} = 14 - 5.94 = 8.06 \][/tex]
So, the pH after adding 125.0 mL of [tex]\(C_6H_5NH_2\)[/tex] is approximately [tex]\(8.06\)[/tex].
### Step 1: Calculate moles of HCl and aniline
Given:
- Volume of HCl ([tex]\(V_{\text{HCl}}\)[/tex]) = 45.0 mL
- Concentration of HCl ([tex]\(C_{\text{HCl}}\)[/tex]) = 0.30 M
- Volume of aniline ([tex]\(V_{\text{aniline}}\)[/tex]) = 125.0 mL
- Concentration of aniline ([tex]\(C_{\text{aniline}}\)[/tex]) = 0.15 M
Convert the volumes from mL to liters (since concentration is in M, moles/L):
[tex]\[ V_{\text{HCl}} = 45.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.045 \, \text{L} \][/tex]
[tex]\[ V_{\text{aniline}} = 125.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.125 \, \text{L} \][/tex]
Calculate moles of HCl:
[tex]\[ \text{moles HCl} = C_{\text{HCl}} \times V_{\text{HCl}} = 0.30 \, \text{M} \times 0.045 \, \text{L} = 0.0135 \, \text{moles} \][/tex]
Calculate moles of aniline:
[tex]\[ \text{moles aniline} = C_{\text{aniline}} \times V_{\text{aniline}} = 0.15 \, \text{M} \times 0.125 \, \text{L} = 0.01875 \, \text{moles} \][/tex]
### Step 2: Determine the limiting reagent and remaining moles
Reaction between HCl and aniline:
[tex]\[ \text{HCl} + \text{C}_6\text{H}_5\text{NH}_2 \rightarrow \text{C}_6\text{H}_5\text{NH}_3^+ + \text{Cl}^- \][/tex]
- Moles of HCl = 0.0135
- Moles of aniline = 0.01875
Since 1 mole of HCl reacts with 1 mole of aniline, the HCl will be the limiting reagent. Calculate remaining moles of aniline:
[tex]\[ \text{moles aniline remaining} = \text{moles aniline} - \text{moles HCl} = 0.01875 - 0.0135 = 0.00525 \, \text{moles} \][/tex]
### Step 3: Calculate the final concentrations
Total volume of the solution:
[tex]\[ V_{\text{total}} = V_{\text{HCl}} + V_{\text{aniline}} = 0.045 \, \text{L} + 0.125 \, \text{L} = 0.170 \, \text{L} \][/tex]
Final concentration of aniline:
[tex]\[ C_{\text{aniline remaining}} = \frac{\text{moles aniline remaining}}{V_{\text{total}}} = \frac{0.00525 \, \text{moles}}{0.170 \, \text{L}} = 0.0309 \, \text{M} \][/tex]
### Step 4: Calculate pH
The remaining aniline ([tex]\(C_6H_5NH_2\)[/tex]) is a weak base. Use the [tex]\(K_b\)[/tex] of aniline to determine the concentration of OH⁻ ions.
Given:
[tex]\[ K_b = 4.3 \times 10^{-10} \][/tex]
For the base dissociation of aniline:
[tex]\[ C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^- \][/tex]
Using the ICE (Initial, Change, Equilibrium) table and simplifying (since [tex]\(x \ll 0.0309\)[/tex]):
[tex]\[ K_b = \frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]} \approx \frac{x^2}{0.0309} \][/tex]
[tex]\[ x^2 = K_b \times 0.0309 \][/tex]
[tex]\[ x^2 = 4.3 \times 10^{-10} \times 0.0309 \][/tex]
[tex]\[ x^2 = 1.3287 \times 10^{-11} \][/tex]
[tex]\[ x \approx \sqrt{1.3287 \times 10^{-11}} \approx 1.152 \times 10^{-6} \][/tex]
Thus,
[tex]\[ [OH^-] = 1.152 \times 10^{-6} \, \text{M} \][/tex]
Calculate pOH:
[tex]\[ pOH = -\log [OH^-] = -\log(1.152 \times 10^{-6}) \approx 5.94 \][/tex]
Finally, calculate pH:
[tex]\[ \text{pH} = 14 - \text{pOH} = 14 - 5.94 = 8.06 \][/tex]
So, the pH after adding 125.0 mL of [tex]\(C_6H_5NH_2\)[/tex] is approximately [tex]\(8.06\)[/tex].
