Answer :
To determine the volume of HCl required and the pH of the solution at the equivalence point, we will follow a systematic approach.
### Step 1: Calculate the Moles of NH₃
We start by calculating the moles of NH₃ we have in the given solution.
Given:
- Concentration of NH₃ (C₁) = 0.35 M
- Volume of NH₃ (V₁) = 12.5 mL
First, convert the volume from mL to L (since molarity is moles/L):
[tex]\[ V_1 = \frac{12.5}{1000} \; \text{L} = 0.0125 \; \text{L} \][/tex]
Now, calculate the moles of NH₃:
[tex]\[ \text{Moles of NH₃} = C_1 \times V_1 \][/tex]
[tex]\[ \text{Moles of NH₃} = 0.35 \times 0.0125 \][/tex]
[tex]\[ \text{Moles of NH₃} = 0.004375 \; \text{moles} \][/tex]
### Step 2: Determine the Volume of HCl required at the Equivalence Point
At the equivalence point, the moles of HCl needed will be equal to the moles of NH₃.
Given:
- Concentration of HCl (C₂) = 0.75 M
We need to find the volume of HCl (V₂) required:
[tex]\[ \text{Moles of HCl} = \text{Moles of NH₃} = 0.004375 \; \text{moles} \][/tex]
Using the molarity formula, we solve for the volume V₂:
[tex]\[ V_2 = \frac{\text{Moles of HCl}}{C_2} \][/tex]
[tex]\[ V_2 = \frac{0.004375}{0.75} \][/tex]
[tex]\[ V_2 = 0.0058333 \; \text{L} \][/tex]
Convert the volume from L to mL:
[tex]\[ V_2 = 0.0058333 \times 1000 \][/tex]
[tex]\[ V_2 = 5.8333 \; \text{mL} \][/tex]
Therefore, the volume of HCl required is [tex]\(5.8333 \; \text{mL}\)[/tex].
### Step 3: Calculate the pH at the Equivalence Point
At the equivalence point, all NH₃ has reacted and has been converted to NH₄⁺. NH₄⁺ is a weak acid, and we need to determine its effect on the pH of the solution.
Given:
- [tex]\( K_b \)[/tex] for NH₃ = [tex]\( 1.79 \times 10^{-5} \)[/tex]
- Kw (ion product of water) = [tex]\( 1 \times 10^{-14} \)[/tex]
First, find the [tex]\( K_a \)[/tex] for NH₄⁺ using the relation:
[tex]\[ K_a = \frac{K_w}{K_b} \][/tex]
[tex]\[ K_a = \frac{1 \times 10^{-14}}{1.79 \times 10^{-5}} \][/tex]
[tex]\[ K_a = 5.586592 \times 10^{-10} \][/tex]
At the equivalence point, the concentration of NH₄⁺ is the same as the initial concentration of NH₃ in the solution because the volume change is relatively small.
The concentration of NH₄⁺:
[tex]\[ \text{Concentration of NH₄⁺} = 0.35 \; \text{M} \][/tex]
To find the pH, we need to determine the [tex]\( [H^+] \)[/tex] concentration. Considering the hydrolysis of NH₄⁺:
[tex]\[ \text{NH}_4^+ \leftrightarrow \text{NH}_3 + H^+ \][/tex]
We apply the expression for [tex]\( K_a \)[/tex]:
[tex]\[ K_a = \frac{[H^+][NH_3]}{[NH_4^+]} \][/tex]
[tex]\[ 5.586592 \times 10^{-10} = \frac{(H^+)^2}{0.35 - H^+} \][/tex]
Assuming the change in concentration [tex]\( \Delta = H^+ \)[/tex] is very small compared to the initial concentration, we simplify:
[tex]\[ (H^+)^2 \approx 5.586592 \times 10^{-10} \times 0.35 \][/tex]
[tex]\[ (H^+)^2 \approx 1.9553072 \times 10^{-10} \][/tex]
Solving for [tex]\( [H^+] \)[/tex]:
[tex]\[ H^+ \approx \sqrt{1.9553072 \times 10^{-10}} \][/tex]
[tex]\[ H^+ \approx 1.3982951 \times 10^{-5} \][/tex]
Finally, calculate the pH:
[tex]\[ \text{pH} = -\log_{10}(H^+) \][/tex]
[tex]\[ \text{pH} = -\log_{10}(1.3982951 \times 10^{-5}) \][/tex]
[tex]\[ \text{pH} \approx 4.8544 \][/tex]
### Summary
- The volume of HCl required is [tex]\(5.8333 \; \text{mL}\)[/tex].
- The pH of the solution at the equivalence point is approximately 4.8544.
### Step 1: Calculate the Moles of NH₃
We start by calculating the moles of NH₃ we have in the given solution.
Given:
- Concentration of NH₃ (C₁) = 0.35 M
- Volume of NH₃ (V₁) = 12.5 mL
First, convert the volume from mL to L (since molarity is moles/L):
[tex]\[ V_1 = \frac{12.5}{1000} \; \text{L} = 0.0125 \; \text{L} \][/tex]
Now, calculate the moles of NH₃:
[tex]\[ \text{Moles of NH₃} = C_1 \times V_1 \][/tex]
[tex]\[ \text{Moles of NH₃} = 0.35 \times 0.0125 \][/tex]
[tex]\[ \text{Moles of NH₃} = 0.004375 \; \text{moles} \][/tex]
### Step 2: Determine the Volume of HCl required at the Equivalence Point
At the equivalence point, the moles of HCl needed will be equal to the moles of NH₃.
Given:
- Concentration of HCl (C₂) = 0.75 M
We need to find the volume of HCl (V₂) required:
[tex]\[ \text{Moles of HCl} = \text{Moles of NH₃} = 0.004375 \; \text{moles} \][/tex]
Using the molarity formula, we solve for the volume V₂:
[tex]\[ V_2 = \frac{\text{Moles of HCl}}{C_2} \][/tex]
[tex]\[ V_2 = \frac{0.004375}{0.75} \][/tex]
[tex]\[ V_2 = 0.0058333 \; \text{L} \][/tex]
Convert the volume from L to mL:
[tex]\[ V_2 = 0.0058333 \times 1000 \][/tex]
[tex]\[ V_2 = 5.8333 \; \text{mL} \][/tex]
Therefore, the volume of HCl required is [tex]\(5.8333 \; \text{mL}\)[/tex].
### Step 3: Calculate the pH at the Equivalence Point
At the equivalence point, all NH₃ has reacted and has been converted to NH₄⁺. NH₄⁺ is a weak acid, and we need to determine its effect on the pH of the solution.
Given:
- [tex]\( K_b \)[/tex] for NH₃ = [tex]\( 1.79 \times 10^{-5} \)[/tex]
- Kw (ion product of water) = [tex]\( 1 \times 10^{-14} \)[/tex]
First, find the [tex]\( K_a \)[/tex] for NH₄⁺ using the relation:
[tex]\[ K_a = \frac{K_w}{K_b} \][/tex]
[tex]\[ K_a = \frac{1 \times 10^{-14}}{1.79 \times 10^{-5}} \][/tex]
[tex]\[ K_a = 5.586592 \times 10^{-10} \][/tex]
At the equivalence point, the concentration of NH₄⁺ is the same as the initial concentration of NH₃ in the solution because the volume change is relatively small.
The concentration of NH₄⁺:
[tex]\[ \text{Concentration of NH₄⁺} = 0.35 \; \text{M} \][/tex]
To find the pH, we need to determine the [tex]\( [H^+] \)[/tex] concentration. Considering the hydrolysis of NH₄⁺:
[tex]\[ \text{NH}_4^+ \leftrightarrow \text{NH}_3 + H^+ \][/tex]
We apply the expression for [tex]\( K_a \)[/tex]:
[tex]\[ K_a = \frac{[H^+][NH_3]}{[NH_4^+]} \][/tex]
[tex]\[ 5.586592 \times 10^{-10} = \frac{(H^+)^2}{0.35 - H^+} \][/tex]
Assuming the change in concentration [tex]\( \Delta = H^+ \)[/tex] is very small compared to the initial concentration, we simplify:
[tex]\[ (H^+)^2 \approx 5.586592 \times 10^{-10} \times 0.35 \][/tex]
[tex]\[ (H^+)^2 \approx 1.9553072 \times 10^{-10} \][/tex]
Solving for [tex]\( [H^+] \)[/tex]:
[tex]\[ H^+ \approx \sqrt{1.9553072 \times 10^{-10}} \][/tex]
[tex]\[ H^+ \approx 1.3982951 \times 10^{-5} \][/tex]
Finally, calculate the pH:
[tex]\[ \text{pH} = -\log_{10}(H^+) \][/tex]
[tex]\[ \text{pH} = -\log_{10}(1.3982951 \times 10^{-5}) \][/tex]
[tex]\[ \text{pH} \approx 4.8544 \][/tex]
### Summary
- The volume of HCl required is [tex]\(5.8333 \; \text{mL}\)[/tex].
- The pH of the solution at the equivalence point is approximately 4.8544.
