Question #20274

1 Answer
Jan 15, 2018

Answer:

#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))#