If the pH of a 0.200 M solution of the CICH2O¯ ion is 12.05, then what is the value for the basicity constant, Kp, of CICH2O-? You do not need to make a simplifying assumption CICH2O¬(aq) + H2O(1) CICH2OH(aq) + OH-(aq) Ky = ?
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![**Title: Calculating the Basicity Constant \( K_b \) of ClCH\(_2\)O\(^-\) Ion**
**Introduction:**
- **Problem Statement:** Given the pH of a 0.200 M solution of ClCH\(_2\)O\(^-\) ion is 12.05, determine the basicity constant, \( K_b \), for ClCH\(_2\)O\(^-\). Note that a simplifying assumption is not required.
**Chemical Equation:**
- **Equilibrium Reaction:**
\[\text{ClCH}_2\text{O}^-(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{ClCH}_2\text{OH}(\text{aq}) + \text{OH}^-(\text{aq})\]
**Goal:**
- **Determine:**
\( K_b = ? \)
**Concepts to Explore:**
1. **Understanding pH and pOH:**
- Given pH, calculate pOH using the formula:
\[\text{pOH} = 14 - \text{pH}\]
2. **Concentration of OH\(^-\):**
- Find the concentration of hydroxide ions, [OH\(^-\)], using:
\[[\text{OH}^-] = 10^{-\text{pOH}}\]
3. **Equilibrium Calculation:**
- Use the equilibrium expression for the reaction:
\[K_b = \frac{[\text{ClCH}_2\text{OH}][\text{OH}^-]}{[\text{ClCH}_2\text{O}^-]}\]
**Steps to Solve:**
1. Calculate the pOH from the given pH.
2. Find the concentration of OH\(^-\) ions.
3. Substitute the equilibrium concentrations into the expression to find \( K_b \).
**Concluding Remarks:**
This exercise demonstrates the application of fundamental equilibrium concepts to calculate the basicity constant of an ion in solution, utilizing given details like pH and molarity.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F092bd7db-9bdb-4d81-a056-eb65c4616336%2F86dac251-5900-4391-bf6c-01941a7849ef%2F2ih7mwh_processed.png&w=3840&q=75)
Given: pH = 12.05
Thus, pOH = 14 - 12.05 = 1.95
As pOH = -log[OH-]
Thus, [OH-] = 10-pOH = 10-1.95 = 0.01122 M
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