A solution has a pOH = 3.45. Calculate [H] [Select] [ Select] 10.55 Calculate [OH 2.8x10-11 2.8x10-3 3.55x10-4

Fundamentals Of Analytical Chemistry
9th Edition
ISBN:9781285640686
Author:Skoog
Publisher:Skoog
Chapter9: Aqueous Solutions And Chemical Equilibria
Section: Chapter Questions
Problem 9.5QAP
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### Problem Statement

A solution has a pOH = 3.45. 

1. **Calculate \([H^+]\)**
   - [Select]

2. **Calculate \([OH^-]\)**
   - [Select]
   - Options provided:
     - 10.55
     - \(2.8 \times 10^{-11}\)
     - \(2.8 \times 10^{-3}\)
     - \(3.55 \times 10^{-4}\) (selected)

### Instructions

To determine the concentrations of \([H^+]\) and \([OH^-]\) in the solution, use the given pOH value and apply the following formulas:

1. **To find \([OH^-]\):**
   \[
   [\text{OH}^-] = 10^{-\text{pOH}}
   \]
   Given \(\text{pOH} = 3.45\), calculate:

   \[
   [\text{OH}^-] = 10^{-3.45}
   \]
2. **To find \([H^+]\):**
   Use the water dissociation constant \(\text{Kw}\):

   \[
   \text{Kw} = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}
   \]

   Thus,

   \[
   [\text{H}^+] = \frac{\text{Kw}}{[\text{OH}^-]} = \frac{1 \times 10^{-14}}{10^{-\text{pOH}}}
   \]

### Multiple Choice Options

- For \([OH^-]\):
  - 10.55
  - \(2.8 \times 10^{-11}\)
  - \(2.8 \times 10^{-3}\)
  - \(3.55 \times 10^{-4}\)

Based on the calculations, please select the correct answer from the dropdown provided.
Transcribed Image Text:### Problem Statement A solution has a pOH = 3.45. 1. **Calculate \([H^+]\)** - [Select] 2. **Calculate \([OH^-]\)** - [Select] - Options provided: - 10.55 - \(2.8 \times 10^{-11}\) - \(2.8 \times 10^{-3}\) - \(3.55 \times 10^{-4}\) (selected) ### Instructions To determine the concentrations of \([H^+]\) and \([OH^-]\) in the solution, use the given pOH value and apply the following formulas: 1. **To find \([OH^-]\):** \[ [\text{OH}^-] = 10^{-\text{pOH}} \] Given \(\text{pOH} = 3.45\), calculate: \[ [\text{OH}^-] = 10^{-3.45} \] 2. **To find \([H^+]\):** Use the water dissociation constant \(\text{Kw}\): \[ \text{Kw} = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14} \] Thus, \[ [\text{H}^+] = \frac{\text{Kw}}{[\text{OH}^-]} = \frac{1 \times 10^{-14}}{10^{-\text{pOH}}} \] ### Multiple Choice Options - For \([OH^-]\): - 10.55 - \(2.8 \times 10^{-11}\) - \(2.8 \times 10^{-3}\) - \(3.55 \times 10^{-4}\) Based on the calculations, please select the correct answer from the dropdown provided.
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