Question
Calculate the pH of the buffer solution prepared by mixing 20 mL of 2 M HNO3 with 50 mL of 0.2 M NaHCO3 solution, given Ka1= 7.25e^-4 and Ka2= 4.28e^-11 for H2CO3.
Answers
GPT-4o mini
To calculate the pH of the buffer solution formed by mixing nitric acid (HNO₃) with sodium bicarbonate (NaHCO₃), we need to consider the components of the system:
1. **HNO₃** is a strong acid and will dissociate completely in solution, contributing protons (H⁺).
2. **NaHCO₃** can act as a weak acid and a weak base in the bicarbonate buffering system. It's weakly basic as it can neutralize some of the added H⁺ and can also dissociate to form H₂CO₃.
### Step 1: Calculate moles of HNO₃
- Volume of HNO₃ solution = 20 mL = 0.020 L
- Concentration of HNO₃ = 2 M
- Moles of HNO₃:
\[
\text{moles of HNO₃} = \text{volume} \times \text{concentration} = 0.020 \, \text{L} \times 2 \, \text{M} = 0.040 \, \text{mol}
\]
### Step 2: Calculate moles of NaHCO₃
- Volume of NaHCO₃ solution = 50 mL = 0.050 L
- Concentration of NaHCO₃ = 0.2 M
- Moles of NaHCO₃:
\[
\text{moles of NaHCO₃} = \text{volume} \times \text{concentration} = 0.050 \, \text{L} \times 0.2 \, \text{M} = 0.010 \, \text{mol}
\]
### Step 3: Determine the interaction between HNO₃ and NaHCO₃
NaHCO₃ can neutralize some of the H⁺ from HNO₃, resulting in the formation of H₂CO₃:
\[
\text{NaHCO₃} + \text{H⁺} \rightarrow \text{H₂CO₃} + \text{Na⁺}
\]
We have:
- 0.040 mol HNO₃ (providing 0.040 mol H⁺)
- 0.010 mol NaHCO₃ (neutralizing 0.010 mol H⁺)
### Step 4: Determine remaining moles after reaction
After neutralization:
- Moles of H⁺ remaining = 0.040 mol - 0.010 mol = 0.030 mol H⁺
- Moles of H₂CO₃ formed = 0.010 mol
- Moles of NaHCO₃ left = 0 (because all 0.010 mol NaHCO₃ was used)
Now, we have:
- \( [H⁺] = 0.030 \, \text{mol} \) (from HNO₃)
- \( [H₂CO₃] = 0.010 \, \text{mol} \)
### Step 5: Calculate concentrations
Total volume after mixing:
\[
V_{\text{total}} = 20 \, \text{mL} + 50 \, \text{mL} = 70 \, \text{mL} = 0.070 \, \text{L}
\]
Concentrations in the final solution:
- Concentration of H⁺:
\[
[H⁺] = \frac{0.030 \, \text{mol}}{0.070 \, \text{L}} \approx 0.429 \, \text{M}
\]
- Concentration of H₂CO₃:
\[
[H₂CO₃] = \frac{0.010 \, \text{mol}}{0.070 \, \text{L}} \approx 0.143 \, \text{M}
\]
### Step 6: Calculate the pH
Using the equilibrium expression for the dissociation of H₂CO₃:
\[
H₂CO₃ \leftrightarrow H⁺ + HCO₃⁻
\]
Using the second dissociation constant \( K_a2 \) for the equilibrium:
\[
K_a2 = \frac{[H^+][HCO₃^-]}{[H₂CO₃]}
\]
Where:
- \( [HCO₃⁻] \) can be assumed to be close to zero initially because all NaHCO₃ has reacted.
- Let's denote \( x = [HCO₃⁻] \) after some is produced.
- So ...
\[
K_a2 = \frac{(0.429)(x)}{0.143}
\]
Setting \( K_a2 \) equal to \( 4.28 \times 10^{-11} \):
\[
4.28 \times 10^{-11} = \frac{(0.429)(x)}{0.143}
\]
Solving for \( x \):
\[
x = \frac{(4.28 \times 10^{-11}) \times 0.143}{0.429} \approx 2.10 \times 10^{-11}
\]
Thus, we can note that \( HCO₃^-\) is effectively negligible compared to \( H^+ \), thus pH can be approximated with \([H^+]\) directly.
Finally:
\[
\text{pH} = -\log[H^+] \approx -\log(0.429) \approx 0.368
\]
### Final answer:
The pH of the solution is approximately **0.37**.
1. **HNO₃** is a strong acid and will dissociate completely in solution, contributing protons (H⁺).
2. **NaHCO₃** can act as a weak acid and a weak base in the bicarbonate buffering system. It's weakly basic as it can neutralize some of the added H⁺ and can also dissociate to form H₂CO₃.
### Step 1: Calculate moles of HNO₃
- Volume of HNO₃ solution = 20 mL = 0.020 L
- Concentration of HNO₃ = 2 M
- Moles of HNO₃:
\[
\text{moles of HNO₃} = \text{volume} \times \text{concentration} = 0.020 \, \text{L} \times 2 \, \text{M} = 0.040 \, \text{mol}
\]
### Step 2: Calculate moles of NaHCO₃
- Volume of NaHCO₃ solution = 50 mL = 0.050 L
- Concentration of NaHCO₃ = 0.2 M
- Moles of NaHCO₃:
\[
\text{moles of NaHCO₃} = \text{volume} \times \text{concentration} = 0.050 \, \text{L} \times 0.2 \, \text{M} = 0.010 \, \text{mol}
\]
### Step 3: Determine the interaction between HNO₃ and NaHCO₃
NaHCO₃ can neutralize some of the H⁺ from HNO₃, resulting in the formation of H₂CO₃:
\[
\text{NaHCO₃} + \text{H⁺} \rightarrow \text{H₂CO₃} + \text{Na⁺}
\]
We have:
- 0.040 mol HNO₃ (providing 0.040 mol H⁺)
- 0.010 mol NaHCO₃ (neutralizing 0.010 mol H⁺)
### Step 4: Determine remaining moles after reaction
After neutralization:
- Moles of H⁺ remaining = 0.040 mol - 0.010 mol = 0.030 mol H⁺
- Moles of H₂CO₃ formed = 0.010 mol
- Moles of NaHCO₃ left = 0 (because all 0.010 mol NaHCO₃ was used)
Now, we have:
- \( [H⁺] = 0.030 \, \text{mol} \) (from HNO₃)
- \( [H₂CO₃] = 0.010 \, \text{mol} \)
### Step 5: Calculate concentrations
Total volume after mixing:
\[
V_{\text{total}} = 20 \, \text{mL} + 50 \, \text{mL} = 70 \, \text{mL} = 0.070 \, \text{L}
\]
Concentrations in the final solution:
- Concentration of H⁺:
\[
[H⁺] = \frac{0.030 \, \text{mol}}{0.070 \, \text{L}} \approx 0.429 \, \text{M}
\]
- Concentration of H₂CO₃:
\[
[H₂CO₃] = \frac{0.010 \, \text{mol}}{0.070 \, \text{L}} \approx 0.143 \, \text{M}
\]
### Step 6: Calculate the pH
Using the equilibrium expression for the dissociation of H₂CO₃:
\[
H₂CO₃ \leftrightarrow H⁺ + HCO₃⁻
\]
Using the second dissociation constant \( K_a2 \) for the equilibrium:
\[
K_a2 = \frac{[H^+][HCO₃^-]}{[H₂CO₃]}
\]
Where:
- \( [HCO₃⁻] \) can be assumed to be close to zero initially because all NaHCO₃ has reacted.
- Let's denote \( x = [HCO₃⁻] \) after some is produced.
- So ...
\[
K_a2 = \frac{(0.429)(x)}{0.143}
\]
Setting \( K_a2 \) equal to \( 4.28 \times 10^{-11} \):
\[
4.28 \times 10^{-11} = \frac{(0.429)(x)}{0.143}
\]
Solving for \( x \):
\[
x = \frac{(4.28 \times 10^{-11}) \times 0.143}{0.429} \approx 2.10 \times 10^{-11}
\]
Thus, we can note that \( HCO₃^-\) is effectively negligible compared to \( H^+ \), thus pH can be approximated with \([H^+]\) directly.
Finally:
\[
\text{pH} = -\log[H^+] \approx -\log(0.429) \approx 0.368
\]
### Final answer:
The pH of the solution is approximately **0.37**.