# A solution of weak base has pH = 7.95. What are the concentrations of the weak base and its conjugate acid ion in this solution? Ka = 6.28*10(-5). BOH <-------> B(+) + OH(-)

Feb 25, 2015

Start by calculating the concentration of the hydroxide ions by using the given value for pH.

$p {H}_{\text{sol}} = 14 - p O H$

$p O H = 14 - p {H}_{\text{sol}} = 14 - 7.95 = 6.05$

$\left[O {H}^{-}\right] = {10}^{- p O H} = {10}^{- 6.05} = 8.91 \cdot {10}^{- 7}$

Now use the equilibrium equation to solve for the initial concentration of the base, which I'll call Z; this in turn will get you the concentration of the base at equilibrium. Set up an ICE chart for your reaction

...$B O H r i g h t \le f t h a r p \infty n s {B}^{+} + O {H}^{-}$
I......Z..............0.............0
C...(-x)............(+x)........(+x)
E...Z-x..............x............x

Don't forget to use the base dissociation constant, ${K}_{b}$, which is equal to

${K}_{b} = {K}_{w} / {K}_{a} = {10}^{- 14} / \left(6.28 \cdot {10}^{- 5}\right) = 1.59 \cdot {10}^{- 10}$

The expression for ${K}_{b}$ is

${K}_{b} = \frac{\left[O {H}^{-}\right] \cdot \left[{B}^{+}\right]}{\left[B O H\right]} = \frac{x \cdot x}{Z - x} = 1.59 \cdot {10}^{- 10}$

Since you know that the value of $x$, the concentration of the hydroxide ions, is equal to $8.91 \cdot {10}^{- 7}$, you can plug this into the above equation to solve for Z

1.59 * 10^(-10) = (8.91 * 10^(-7))^(2)/(Z - 8.91 * 10^(-7)

$Z = {\left(8.91 \cdot {10}^{- 7}\right)}^{2} / \left(1.59 \cdot {10}^{- 10}\right) + 8.91 \cdot {10}^{- 7} = 4.99 \cdot {10}^{- 3} + 8.91 \cdot {10}^{- 7}$

It's same to assume that Z will be $Z \cong 4.99 \cdot {10}^{- 3}$. Because the base dissociation constant is so small, the change in the concentration of the base can be considered unimportant.

This means that the concentration of the weak base and of its conjugate acid in solution will be

$\left[B O H\right] = 4.99 \cdot {10}^{- 3} - 8.91 \cdot {10}^{- 7} \cong 4.99 \cdot {10}^{- 3} \text{M}$
$\left[{B}^{+}\right] = \left[O {H}^{-}\right] = 8.91 \cdot {10}^{- 7} \text{M}$

SIDE NOTE. I'm assuming these concentrations to be in mol/L.