## Chemistry and Chemical Reactivity (9th Edition)

$[Base] = 0.142M$ $[OH^-] = [Conj. Acid] = 8.414 \times 10^{- 3}M$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [Conj.Acid] = x$ -$[Base] = [Base]_{initial} - x = 0.15 - x$ For approximation, we consider: $[Base] = 0.15M$ 2. Now, use the Kb value and equation to find the 'x' value. $Ka = \frac{[OH^-][Conj.Acid]}{ [Base]}$ $Ka = 5 \times 10^{- 4}= \frac{x * x}{ 0.15}$ $Ka = 5 \times 10^{- 4}= \frac{x^2}{ 0.15}$ $7.5 \times 10^{- 5} = x^2$ $x = 8.66 \times 10^{- 3}$ Percent ionization: $\frac{ 8.66 \times 10^{- 3}}{ 0.15} \times 100\% = 5.774\%$ %ionization <>5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration: $Kb = 5 \times 10^{- 4}= \frac{x^2}{ 0.15- x}$ $7.5 \times 10^{- 5} - 5 \times 10^{- 4}x = x^2$ $7.5 \times 10^{- 5} - 5 \times 10^{- 4}x - x^2 = 0$ Bhaskara: $\Delta = (- 5 \times 10^{- 4})^2 - 4 * (-1) *( 7.5 \times 10^{- 5})$ $\Delta = 2.5 \times 10^{- 7} + 3 \times 10^{- 4} = 3.002 \times 10^{- 4}$ $x_1 = \frac{ - (- 5 \times 10^{- 4})+ \sqrt { 3.002 \times 10^{- 4}}}{2*(-1)}$ or $x_2 = \frac{ - (- 5 \times 10^{- 4})- \sqrt { 3.002 \times 10^{- 4}}}{2*(-1)}$ $x_1 = - 8.914 \times 10^{- 3} (Negative)$ $x_2 = 8.414 \times 10^{- 3}$ The concentration can't be negative. Therefore: $[OH^-] = [Conj. Acid] = x = 8.414 \times 10^{- 3}M$ And: $[Base] = 0.15 - x = 0.15 - 8.414 \times 10^{-3} = 0.142M$