Question
Calculate the ratio of H+ ions to OH– ions at a pH = 8. Find the concentration of H+ ions to OH– ions listed in Table B of your Student Guide. Then divide the H+ concentration by the OH– concentration. Record this calculated ratio in Table A of your Student Guide.
Compare your approximated and calculated ratios of H+ ions to OH– ions at a pH = 8. Are they the same? Why or why not? Record your explanation in Table A.
What is the concentration of H+ ions at a pH = 8?
mol/L
What is the concentration of OH– ions at a pH = 8?
mol/L
What is the ratio of H+ ions to OH– ions at a pH = 8?
:1 OR 1:
Compare your approximated and calculated ratios of H+ ions to OH– ions at a pH = 8. Are they the same? Why or why not? Record your explanation in Table A.
What is the concentration of H+ ions at a pH = 8?
mol/L
What is the concentration of OH– ions at a pH = 8?
mol/L
What is the ratio of H+ ions to OH– ions at a pH = 8?
:1 OR 1:
Answers
GPT-4o mini
To calculate the concentrations of H⁺ and OH⁻ ions at a pH of 8, we can use the following formulas:
1. The concentration of H⁺ ions can be calculated from the pH using the formula:
\[
[H^+] = 10^{-\text{pH}}
\]
For a pH of 8:
\[
[H^+] = 10^{-8} \, \text{mol/L} = 1.0 \times 10^{-8} \, \text{mol/L}
\]
2. The concentration of OH⁻ ions can be calculated using the relationship between pH and pOH:
\[
\text{pOH} = 14 - \text{pH}
\]
So, at a pH of 8:
\[
\text{pOH} = 14 - 8 = 6
\]
We can now calculate the concentration of OH⁻:
\[
[OH^-] = 10^{-\text{pOH}} = 10^{-6} \, \text{mol/L} = 1.0 \times 10^{-6} \, \text{mol/L}
\]
Now we can calculate the ratio of H⁺ ions to OH⁻ ions:
\[
\text{Ratio} = \frac{[H^+]}{[OH^-]} = \frac{1.0 \times 10^{-8} \, \text{mol/L}}{1.0 \times 10^{-6} \, \text{mol/L}} = \frac{1}{100} = 1:100
\]
### Summary of Values:
- Concentration of H⁺ ions at pH = 8: **1.0 x 10⁻⁸ mol/L**
- Concentration of OH⁻ ions at pH = 8: **1.0 x 10⁻⁶ mol/L**
- Ratio of H⁺ to OH⁻ at pH = 8: **1:100**
### Explanation:
The approximated and calculated ratios are expected to match, given that the relationships between H⁺ and OH⁻ concentrations are derived from the same principles of acid-base chemistry. However, if there is a discrepancy, it could be due to rounding differences or approximations used during calculation.
For your record in Table A:
- Concentration of H⁺ ions: 1.0 x 10⁻⁸ mol/L
- Concentration of OH⁻ ions: 1.0 x 10⁻⁶ mol/L
- Ratio of H⁺ to OH⁻: 1:100
You will want to confirm if you've referenced any data or approximations that would suggest different values and clarify your findings based on that.
1. The concentration of H⁺ ions can be calculated from the pH using the formula:
\[
[H^+] = 10^{-\text{pH}}
\]
For a pH of 8:
\[
[H^+] = 10^{-8} \, \text{mol/L} = 1.0 \times 10^{-8} \, \text{mol/L}
\]
2. The concentration of OH⁻ ions can be calculated using the relationship between pH and pOH:
\[
\text{pOH} = 14 - \text{pH}
\]
So, at a pH of 8:
\[
\text{pOH} = 14 - 8 = 6
\]
We can now calculate the concentration of OH⁻:
\[
[OH^-] = 10^{-\text{pOH}} = 10^{-6} \, \text{mol/L} = 1.0 \times 10^{-6} \, \text{mol/L}
\]
Now we can calculate the ratio of H⁺ ions to OH⁻ ions:
\[
\text{Ratio} = \frac{[H^+]}{[OH^-]} = \frac{1.0 \times 10^{-8} \, \text{mol/L}}{1.0 \times 10^{-6} \, \text{mol/L}} = \frac{1}{100} = 1:100
\]
### Summary of Values:
- Concentration of H⁺ ions at pH = 8: **1.0 x 10⁻⁸ mol/L**
- Concentration of OH⁻ ions at pH = 8: **1.0 x 10⁻⁶ mol/L**
- Ratio of H⁺ to OH⁻ at pH = 8: **1:100**
### Explanation:
The approximated and calculated ratios are expected to match, given that the relationships between H⁺ and OH⁻ concentrations are derived from the same principles of acid-base chemistry. However, if there is a discrepancy, it could be due to rounding differences or approximations used during calculation.
For your record in Table A:
- Concentration of H⁺ ions: 1.0 x 10⁻⁸ mol/L
- Concentration of OH⁻ ions: 1.0 x 10⁻⁶ mol/L
- Ratio of H⁺ to OH⁻: 1:100
You will want to confirm if you've referenced any data or approximations that would suggest different values and clarify your findings based on that.