Answer
The pH of that $0.010M$ $HB$ solution is equal to $3.66$
Work Step by Step
$Na^+$ 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
$pH = -log[H_3O^+]$
$pH = -log( 2.2 \times 10^{- 4})$
$pH = 3.66$
