For the reaction below, Kc = 1.10 x 10-8. What is the equilibrium concentration of OH- if the reaction begins with 0.420 M HONH2? HONH2 (aq) + H2O (I) = HONH:* (aq) + OH- (aq)

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Chapter1: Chemical Foundations
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**Problem Statement:**

For the reaction below, \( K_c = 1.10 \times 10^{-8} \). What is the equilibrium concentration of \( \text{OH}^- \) if the reaction begins with 0.420 M \( \text{HONH}_2 \)?

**Chemical Equation:**

\[ \text{HONH}_2 \, (\text{aq}) + \text{H}_2\text{O} \, (\text{l}) \rightleftharpoons \text{HONH}_3^+ \, (\text{aq}) + \text{OH}^- \, (\text{aq}) \]

**Explanation:**

This problem involves calculating the equilibrium concentration of hydroxide ions (\( \text{OH}^- \)) for a given reaction mixture. The initial concentration of the reactant \( \text{HONH}_2 \) is provided, and the equilibrium constant (\( K_c \)) is given. The problem requires setting up an expression for \( K_c \) in terms of the concentrations of the products and reactants at equilibrium and solving for the unknown equilibrium concentration of \( \text{OH}^- \).
Transcribed Image Text:**Problem Statement:** For the reaction below, \( K_c = 1.10 \times 10^{-8} \). What is the equilibrium concentration of \( \text{OH}^- \) if the reaction begins with 0.420 M \( \text{HONH}_2 \)? **Chemical Equation:** \[ \text{HONH}_2 \, (\text{aq}) + \text{H}_2\text{O} \, (\text{l}) \rightleftharpoons \text{HONH}_3^+ \, (\text{aq}) + \text{OH}^- \, (\text{aq}) \] **Explanation:** This problem involves calculating the equilibrium concentration of hydroxide ions (\( \text{OH}^- \)) for a given reaction mixture. The initial concentration of the reactant \( \text{HONH}_2 \) is provided, and the equilibrium constant (\( K_c \)) is given. The problem requires setting up an expression for \( K_c \) in terms of the concentrations of the products and reactants at equilibrium and solving for the unknown equilibrium concentration of \( \text{OH}^- \).
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