# Chapter 16 - Applications of Aqueous Equilibria - Worked Example - Page 663: 4

$pH = 10.08$

#### Work Step by Step

1. Drawing the ICE table we get these concentrations at the equilibrium: $CH_3NH_2(aq) + H_2O(l) \lt -- \gt CH_3N{H_3}^+(aq) + OH^-(aq)$ Remember: Reactants at equilibrium = Initial Concentration - x And Products = Initial Concentration + x $[CH_3NH_2] = 0.2 M - x$ $[CH_3N{H_3}^+] = 0.6M + x$ $[OH^-] = 0 + x$ 2. Calculate 'x' using the $K_b$ expression. $3.7\times 10^{- 4} = \frac{[CH_3N{H_3}^+][OH^-]}{[CH_3NH_2]}$ $3.7\times 10^{- 4} = \frac{( 0.6 + x )* x}{ 0.2 - x}$ Considering 'x' has a very small value. $3.7\times 10^{- 4} = \frac{ 0.6 * x}{ 0.2}$ $3.7\times 10^{- 4} = 3x$ $\frac{ 3.7\times 10^{- 4}}{ 3} = x$ $x = 1.2\times 10^{- 4}$ Percent ionization: $\frac{ 1.2\times 10^{- 4}}{ 0.2} \times 100\% = 0.062\%$ x = $[OH^-]$ 3. Calculate the pOH: $pOH = -log[OH^-]$ $pOH = -log( 1.2 \times 10^{- 4})$ $pOH = 3.92$ 4. Find the pH: $pH + pOH = 14$ $pH + 3.92 = 14$ $pH = 10.08$

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