Determine the H*|, pH, and pOH of a solution with an OH-] of 4.2 x 10-12 M at 25 °C. [H*] = M %3D pH = pOH = Determine the H*], OH-, and pOH of a solution with a pH of 10.33 at 25 °C.
Determine the H*|, pH, and pOH of a solution with an OH-] of 4.2 x 10-12 M at 25 °C. [H*] = M %3D pH = pOH = Determine the H*], OH-, and pOH of a solution with a pH of 10.33 at 25 °C.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Chemistry Calculations: Understanding pH and pOH**
1. **Determine the \([H^+]\), pH, and pOH of a solution with an \([OH^-]\) of \(4.2 \times 10^{-12}\) M at 25°C.**
- \([H^+]\) = __________ M
- pH = __________
- pOH = __________
2. **Determine the \([H^+]\), \([OH^-]\), and pOH of a solution with a pH of 10.33 at 25°C.**
**Instructions:**
To solve these problems, use the following formulas and relationships:
- \([H^+]\) and \([OH^-]\) are related by the ion product of water: \([H^+][OH^-] = 1.0 \times 10^{-14}\) at 25°C.
- The pH is calculated as: \( \text{pH} = -\log[H^+]\)
- The pOH is calculated as: \( \text{pOH} = -\log[OH^-]\)
- The relationship between pH and pOH is: \( \text{pH} + \text{pOH} = 14\)
Fill in the blanks with the appropriate values calculated using these relationships.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa13eef44-f673-4796-be37-c768f494742c%2F57003ad0-b4f1-4e22-bb89-602e7086f4f2%2Fl936scl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemistry Calculations: Understanding pH and pOH**
1. **Determine the \([H^+]\), pH, and pOH of a solution with an \([OH^-]\) of \(4.2 \times 10^{-12}\) M at 25°C.**
- \([H^+]\) = __________ M
- pH = __________
- pOH = __________
2. **Determine the \([H^+]\), \([OH^-]\), and pOH of a solution with a pH of 10.33 at 25°C.**
**Instructions:**
To solve these problems, use the following formulas and relationships:
- \([H^+]\) and \([OH^-]\) are related by the ion product of water: \([H^+][OH^-] = 1.0 \times 10^{-14}\) at 25°C.
- The pH is calculated as: \( \text{pH} = -\log[H^+]\)
- The pOH is calculated as: \( \text{pOH} = -\log[OH^-]\)
- The relationship between pH and pOH is: \( \text{pH} + \text{pOH} = 14\)
Fill in the blanks with the appropriate values calculated using these relationships.
![**Exercise: Calculating pH, pOH, and Ion Concentrations**
**Problem 1:**
Determine the \([OH^-]\), pH, and pOH of a solution with a \([H^+]\) of \(5.9 \times 10^{-6}\) M at 25 °C.
- \([OH^-]\) = \_\_\_\_\_\_\_\_ M
- pH = \_\_\_\_\_\_\_\_
- pOH = \_\_\_\_\_\_\_\_
---
**Problem 2:**
Determine the \([H^+]\), pH, and pOH of a solution with an \([OH^-]\) of \(4.2 \times 10^{-12}\) M at 25 °C.
- \([H^+]\) = \_\_\_\_\_\_\_\_ M
- pH = \_\_\_\_\_\_\_\_
- pOH = \_\_\_\_\_\_\_\_
---
**Guidance for Solutions:**
- To find \([OH^-]\) when \([H^+]\) is known, use the relationship:
\[
[H^+] \times [OH^-] = 1.0 \times 10^{-14} \, \text{at 25 °C}
\]
- To calculate pH, use:
\[
\text{pH} = -\log[H^+]
\]
- For pOH, the formula is:
\[
\text{pOH} = -\log[OH^-]
\]
- Remember:
\[
\text{pH} + \text{pOH} = 14 \, \text{at 25 °C}
\]
Use these formulas to fill in the blanks and determine the required values.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa13eef44-f673-4796-be37-c768f494742c%2F57003ad0-b4f1-4e22-bb89-602e7086f4f2%2Fcm7mctw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Exercise: Calculating pH, pOH, and Ion Concentrations**
**Problem 1:**
Determine the \([OH^-]\), pH, and pOH of a solution with a \([H^+]\) of \(5.9 \times 10^{-6}\) M at 25 °C.
- \([OH^-]\) = \_\_\_\_\_\_\_\_ M
- pH = \_\_\_\_\_\_\_\_
- pOH = \_\_\_\_\_\_\_\_
---
**Problem 2:**
Determine the \([H^+]\), pH, and pOH of a solution with an \([OH^-]\) of \(4.2 \times 10^{-12}\) M at 25 °C.
- \([H^+]\) = \_\_\_\_\_\_\_\_ M
- pH = \_\_\_\_\_\_\_\_
- pOH = \_\_\_\_\_\_\_\_
---
**Guidance for Solutions:**
- To find \([OH^-]\) when \([H^+]\) is known, use the relationship:
\[
[H^+] \times [OH^-] = 1.0 \times 10^{-14} \, \text{at 25 °C}
\]
- To calculate pH, use:
\[
\text{pH} = -\log[H^+]
\]
- For pOH, the formula is:
\[
\text{pOH} = -\log[OH^-]
\]
- Remember:
\[
\text{pH} + \text{pOH} = 14 \, \text{at 25 °C}
\]
Use these formulas to fill in the blanks and determine the required values.
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