## Chemistry 9th Edition

The pH of that $0.010M$ $HB$ solution is equal to $3.66$
$Na^+$ doesn't affect the pH of the solution. So, we can consider a $0.050M$ $B^-$ solution, and find the $K_b$ for that base using the given pH. 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $B^-(aq) + H_2O(l) \lt -- \gt HB(aq) + OH^-(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $HB$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M -$[OH^-] = [HB] = 0 + x = x$ -$[B^-] = [B^-]_{initial} - x$ For approximation, we are going to consider $[B^-]_{initial} = [B^-]$ 2. Calculate the hydroxide ion concentration: pH + pOH = 14 9.00 + pOH = 14 pOH = 5.00 $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 5.00}$ $[OH^-] = 1.0 \times 10^{- 5}$ Therefore : $x = 1.0 \times 10^{- 5}$ 3. Write the $K_b$ equation, and find its value: $Kb = \frac{[OH^-][HB]}{ [B^-]}$ $Kb = \frac{x^2}{[Initial B^-] - x}$ $Kb = \frac{( 1.0\times 10^{- 5})^2}{ 0.050- 1.0\times 10^{- 5}}$ $Kb = \frac{ 1.0\times 10^{- 10}}{ 0.050}$ $Kb = 2.0 \times 10^{- 9}$ 4. Since $HB$ is the conjugate acid of $B^-$ , we can calculate its $K_a$ by using this equation: $K_b * K_a = K_w = 10^{-14}$ $2.0 \times 10^{- 9} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 2.0\times 10^{- 9}}$ $K_a = 5.0 \times 10^{- 6}$ Now, use the $K_a$ value we have discovered to calculate the pH for that 0.010M $HB$ solution. 5. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $HB(aq) + H_2O(l) \lt -- \gt B^-(aq) + H_3O^+(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $B^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M -$[H_3O^+] = [B^-] = 0 + x = x$ -$[HB] = [HB]_{initial} - x$ For approximation, we are going to consider $[HB]_{initial} = [HB]$ 6. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][B^-]}{ [HB]}$ $Ka = 5.0 \times 10^{- 6}= \frac{x * x}{ 0.010}$ $Ka = 5.0 \times 10^{- 6}= \frac{x^2}{ 0.010}$ $x^2 = 0.010 \times 5.0 \times 10^{-6}$ $x = \sqrt { 0.010 \times 5.0 \times 10^{-6}} = 2.2 \times 10^{-4}$ Percent dissociation: $\frac{ 2.2 \times 10^{- 4}}{ 0.010} \times 100\% = 2.2\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [B^-] = x = 2.2 \times 10^{- 4}M$ 7. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 2.2 \times 10^{- 4})$ $pH = 3.66$