## Chemistry: Atoms First (2nd Edition)

$[OH^-] = [H_2NN{H_3}^+] = 2.449 \times 10^{- 3}M$ $[H_2NNH_2] \approx 2M$ $pH = 11.389$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [H_2NN{H_3}^+] = x$ -$[H_2NNH_2] = [H_2NNH_2]_{initial} - x = 2 - x$ For approximation, we consider: $[H_2NNH_2] = 2M$ 2. Now, use the Kb value and equation to find the 'x' value. $Ka = \frac{[OH^-][H_2NN{H_3}^+]}{ [H_2NNH_2]}$ $Ka = 3 \times 10^{- 6}= \frac{x * x}{ 2}$ $Ka = 3 \times 10^{- 6}= \frac{x^2}{ 2}$ $6 \times 10^{- 6} = x^2$ $x = 2.449 \times 10^{- 3}$ Percent ionization: $\frac{ 2.449 \times 10^{- 3}}{ 2} \times 100\% = 0.1225\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [H_2NN{H_3}^+] = x = 2.449 \times 10^{- 3}M$ $[H_2NNH_2] \approx 2M$ $pOH = -log[OH^-]$ $pOH = -log( 2.449 \times 10^{- 3})$ $pOH = 2.611$ $pH + pOH = 14$ $pH + 2.611 = 14$ $pH = 11.389$