IDNLearn.com offers a comprehensive solution for finding accurate answers quickly. Get the information you need from our community of experts who provide accurate and comprehensive answers to all your questions.
Sagot :
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.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Thank you for choosing IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more solutions.
