# Question 20274

Jan 15, 2018

$14 - \text{p} {K}_{b}$

#### Explanation:

For a buffer that contains equal concentrations of a weak base and of its conjugate acid, the $\text{pH}$ is given by the $\text{p} {K}_{b}$ of the weak base.

(["conjugate acid"])/(["weak base"]) = 1 implies color(blue)(ul(color(black)("pH" = 14 - "p"K_b)))

This is the case because a weak base - conjugate acid buffer that has equal concentrations of weak base and of conjugate acid has

$\text{pOH" = "p} {K}_{b}$

As you know, an aqueous solution at ${25}^{\circ} \text{C}$ has

$\text{pH " + " pOH} = 14$

This means that the buffer has

$14 - \text{pH" = "p} {K}_{b}$

which, of course, gets you

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{pH" = 14 - "p} {K}_{b}}}}$

Alternatively, you can use the fact that at ${25}^{\circ} \text{C}$, an aqueous solution has

$\text{pK"_a + "p} {K}_{b} = 14$

This means that

$\text{p"K_b = 14 - "pH}$

which, once again, gets you

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{pH" = 14 - "p} {K}_{b}}}}$

When in doubt, remember that the $\text{pH}$ of a buffer that contains a weak base and its conjugate acid can be calculated by using this form of the Henderson - Hasselbalch equation

"pOH" = "p"K_b + log( (["conjugate acid"])/(["weak base"]))

This can be rewritten as

14 - "pH" = "p"K_b + log( (["conjugate acid"])/(["weak base"]))

to get

"pH" = 14 - ["p"K_b + log( (["conjugate acid"])/(["weak base"]))]

Notice that when you have

$\left(\left[\text{conjugate acid"])/(["weak base}\right]\right) = 1$

you get

"pH" = 14 - ["p"K_b + log(1)]#

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{pH"= 14 - "p} {K}_{b}}}}$

So, to find the $\text{pH}$ of the solution, use the fact that

$\text{p} {K}_{b} = - \log \left({K}_{b}\right)$

to get

$\text{pH} = 14 - \left[- \log \left(1.8 \cdot {10}^{- 5}\right)\right]$

$\text{pH} = 14 + \log \left(1.8 \cdot {10}^{- 5}\right)$