Question

Question #20274

Answer 1

`14 - "p"K_b`

Explanation:

For a buffer that contains equal concentrations of a weak base and of its conjugate acid, the `"pH"` is given by the `"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

`"pOH" = "p"K_b`

As you know, an aqueous solution at `25^@"C"` has

`"pH " + " pOH" = 14`

This means that the buffer has

`14 - "pH" = "p"K_b`

which, of course, gets you

`color(blue)(ul(color(black)("pH" = 14 - "p"K_b)))`

Alternatively, you can use the fact that at `25^@"C"`, an aqueous solution has

`"pK"_a + "p"K_b = 14`

This means that

`"p"K_b = 14 - "pH"`

which, once again, gets you

`color(blue)(ul(color(black)("pH" = 14 - "p"K_b)))`

When in doubt, remember that the `"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

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

you get

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

`color(blue)(ul(color(black)("pH"= 14 - "p"K_b)))`

So, to find the `"pH"` of the solution, use the fact that

`"p"K_b = - log(K_b)`

to get

`"pH" = 14 - [- log(1.8 * 10^(-5))]`

`"pH" = 14 + log(1.8 * 10^(-5))`