Question b75b4

Jan 14, 2018

$\left(\left[\text{conjugate base"])/(["weak acid}\right]\right) = 10$

Explanation:

You know that when the buffer contains equal concentrations of weak acid and of conjugate base, you have

(["conjugate base"])/(["weak acid"]) = 1 implies "pH" = "p"K_a

You also know that the $\text{pH}$ is calculated using a logarithm scale of base $10$, so if the $\text{pH}$ increases by $1$ unit, then the ratio of the conjugate base to the weak acid must increase by $10$.

(["conjugate base"])/(["weak acid"]) = 10 implies "pH" = "p"K_a + 1

Now, this is the case because the $\text{pH}$ of a buffer that contains a weak acid and its conjugate base can be calculated by using the Henderson - Hasselbalch equation

"pH" = "p"K_a = log( (["conjugate base"])/(["weak acid"]))#

As you can see, the ratio that exists between the concentration of the conjugate base and the concentration of the weak acid determines the difference between the $\text{pH}$ of the buffer and the $\text{p} {K}_{a}$ of the weak acid.

Notice that in order o get the $\text{pH}$ of the buffer to decrease by $1$ unit, you need to have

$\left(\left[\text{conjugate base"])/(["weak acid}\right]\right) = \frac{1}{10}$

since

$\text{pH" = "p} {K}_{a} + \log \left(\frac{1}{10}\right)$

$\text{pH" = "p} {K}_{a} + \left(- 1\right)$

$\text{pH" = "p} {K}_{a} - 1$

So remember, if the concentration of the conjugate base is $10$ times higher than the concentration of the weak acid, the $\text{pH}$ of the buffer is $1$ unit higher than the $\text{p} {K}_{a}$ of the weak acid.

$\left\{\begin{matrix}\text{10 times more conjugate base " implies " pH" = "p"K_a + 1 \\ "10 times more weak acid " implies " pH" = "p} {K}_{a} - 1\end{matrix}\right.$