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The pH of 0.25 M methylamine is 11.98 at 25 °C. The [OH-] at equilibrium is given below. What is the Kb for methylamine? CH3NH2 + H₂O = CH3NH + OH- [OH-] = 9.5 x 10-³ M Kb = [?] x 10¹?] M X Coefficient (green) Exponent (yellow) Enter

The pH of 0.25 M methylamine is 11.98 at 25 °C. The [OH-] at equilibrium is given below. What is the Kb for methylamine? CH3NH2 + H₂O = CH3NH + OH- [OH-] = 9.5 x 10-³ M Kb = [?] x 10¹?] M X Coefficient (green) Exponent (yellow) Enter

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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### Determining the Base Dissociation Constant (\( K_b \)) for Methylamine **Problem Statement:** The pH of a 0.25 M methylamine solution is 11.98 at 25 °C. The hydroxide ion concentration \([OH^-]\) at equilibrium is given below. What is the \( K_b \) for methylamine? **Chemical Equation:** \[ \text{CH}_3 \text{NH}_2 + \text{H}_2 \text{O} \rightleftharpoons \text{CH}_3 \text{NH}_3^+ + \text{OH}^- \] **Given:** \[ [\text{OH}^-] = 9.5 \times 10^{-3} \text{ M} \] **Task:** Calculate the base dissociation constant (\( K_b \)) and express it in the form: \[ K_b = [ \boxed{?} ] \times 10^{\boxed{?}} \text{ M} \] **Input Fields:** - Coefficient (green box) - Exponent (yellow box) **Explanation:** 1. **Write the Expression for the Base Dissociation Constant:** The base dissociation constant \( K_b \) for methylamine can be expressed as: \[ K_b = \frac{[\text{CH}_3 \text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3 \text{NH}_2]} \] 2. **Use the Given Data:** - The concentration of hydroxide ions, \([OH^-]\), is \( 9.5 \times 10^{-3} \) M. - The initial concentration of methylamine, \( [\text{CH}_3 \text{NH}_2] \), is 0.25 M. 3. **Calculations:** - Assume that the change in concentration of \([CH_3NH_2]\) is negligible compared to the initial concentration. - The equilibrium concentration of \([CH_3NH_3^+]\) is roughly equal to \([OH^-]\). - Using these assumptions, calculate the value of \( K_b \). 4. **Result Format:** - Fill in the coefficient in the green box. - Fill in the exponent in the yellow box. **Interactive
Transcribed Image Text:### Determining the Base Dissociation Constant (\( K_b \)) for Methylamine **Problem Statement:** The pH of a 0.25 M methylamine solution is 11.98 at 25 °C. The hydroxide ion concentration \([OH^-]\) at equilibrium is given below. What is the \( K_b \) for methylamine? **Chemical Equation:** \[ \text{CH}_3 \text{NH}_2 + \text{H}_2 \text{O} \rightleftharpoons \text{CH}_3 \text{NH}_3^+ + \text{OH}^- \] **Given:** \[ [\text{OH}^-] = 9.5 \times 10^{-3} \text{ M} \] **Task:** Calculate the base dissociation constant (\( K_b \)) and express it in the form: \[ K_b = [ \boxed{?} ] \times 10^{\boxed{?}} \text{ M} \] **Input Fields:** - Coefficient (green box) - Exponent (yellow box) **Explanation:** 1. **Write the Expression for the Base Dissociation Constant:** The base dissociation constant \( K_b \) for methylamine can be expressed as: \[ K_b = \frac{[\text{CH}_3 \text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3 \text{NH}_2]} \] 2. **Use the Given Data:** - The concentration of hydroxide ions, \([OH^-]\), is \( 9.5 \times 10^{-3} \) M. - The initial concentration of methylamine, \( [\text{CH}_3 \text{NH}_2] \), is 0.25 M. 3. **Calculations:** - Assume that the change in concentration of \([CH_3NH_2]\) is negligible compared to the initial concentration. - The equilibrium concentration of \([CH_3NH_3^+]\) is roughly equal to \([OH^-]\). - Using these assumptions, calculate the value of \( K_b \). 4. **Result Format:** - Fill in the coefficient in the green box. - Fill in the exponent in the yellow box. **Interactive
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