Question #b75b4

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Question #b75b4

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Answer 1 · 2018-01-14T14:51:31

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

Explanation:

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

$\frac{[conjugate base]}{[weak acid]} = 1 \Rightarrow pH = p K_{a}$ $(["conjugate base"])/(["weak acid"]) = 1 implies "pH" = "p"K_a$

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

$\frac{[conjugate base]}{[weak acid]} = 10 \Rightarrow pH = p K_{a} + 1$ $(["conjugate base"])/(["weak acid"]) = 10 implies "pH" = "p"K_a + 1$

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

$pH = p K_{a} = log (\frac{[conjugate base]}{[weak acid]})$ $"pH" = "p"K_a = log( (["conjugate base"])/(["weak acid"]))$

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 $pH$ $"pH"$ of the buffer and the $p K_{a}$ $"p"K_a$ of the weak acid.

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

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

since

$pH = p K_{a} + log (\frac{1}{10})$ $"pH" = "p"K_a + log(1/10)$

$pH = p K_{a} + (- 1)$ $"pH" = "p"K_a + (-1)$

$pH = p K_{a} - 1$ $"pH" = "p"K_a - 1$

So remember, if the concentration of the conjugate base is $10$ $10$ times higher than the concentration of the weak acid, the $pH$ $"pH"$ of the buffer is $1$ $1$ unit higher than the $p K_{a}$ $"p"K_a$ of the weak acid.

${("10 times more conjugate base " implies " pH" = "p"K_a + 1),("10 times more weak acid " implies " pH" = "p"K_a -1) :}$

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Question #b75b4

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