## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Self-Assessment Quiz - Page 768: Q10

(d) 0.011 M

#### Work Step by Step

1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ C_2H_5NH_2 ]& [ C_2H_5{NH_3}^{+} ]& [ OH^- ]\\ Initial& 0.200 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.200 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_b$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_b = \frac{[Products]}{[Reactants]} = \frac{[ C_2H_5{NH_3}^{+} ][ H^+ ]}{[ C_2H_5NH_2 ]}$$ $$K_b = \frac{(x)(x)}{[ C_2H_5NH_2 ]_{initial} - x}$$ 3. Assuming $0.200 \gt\gt x$: $$K_b = \frac{x^2}{[ C_2H_5NH_2 ]_{initial}}$$ $$x = \sqrt{K_b \times [ C_2H_5NH_2 ]_{initial}} = \sqrt{ 5.6 \times 10^{-4} \times 0.200 }$$ $x = 0.011$ 4. Test if the assumption was correct: $$\frac{ 0.011 }{ 0.200 } \times 100\% = 5 \%$$ $[OH^-] = x = 0.011 \space M$

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