Question

Question #b75b4

Answer 1

`(["conjugate base"])/(["weak acid"]) = 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 `"pH"` is calculated using a logarithm scale of base `10`, so if the `"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 `"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 `"pH"` of the buffer and the `"p"K_a` of the weak acid.

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

`(["conjugate base"])/(["weak acid"]) = 1/10`

since

`"pH" = "p"K_a + log(1/10)`

`"pH" = "p"K_a + (-1)`

`"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 `"pH"` of the buffer is `1` unit higher than the `"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) :}`