If the #K_b# of a weak base is #1.2 times 10^-6#, what is the pH of a 0.34 M solution of this base?

1 Answer
Truong-Son N.
May 4, 2017

It does not matter what base you are looking at. If you know its #K_b# and its concentration, that is enough. Denote a neutral base as #B#. Then the ICE table can be constructed.

#"B"(aq) " "+" " "H"_2"O"(l) rightleftharpoons "BH"^(+)(aq) + "OH"^(-)(aq)#

#"I"" ""0.34 M"" "" "" "-" "" "" ""0 M"" "" "" ""0 M"#
#"C"" "-x" "" "" "" "-" "" "+x" "" "" "+x#
#"E"" "(0.34 - x)"M"" "-" "" "" "x" "" "" "" "x#

Thus, the #K_b# is:

#K_b = x^2/(0.34 - x)#

In fact, the #K_b# is small enough to use the small #x# approximation. A good rule of thumb is to check whether #K# is on the order of #10^(-5)#. Hence, we have:

#K_b ~~ x^2/0.34#

and the equilibrium concentration of #"OH"^(-)# is readily obtained:

#x = sqrt(0.34K_b)#

#= 6.39 xx 10^(-4)# #"M"#

Therefore, the #"pH"# is:

#color(blue)("pH") = 14 - "pOH"#

#= 14 - (-log["OH"^(-)])#

#= 14 - 3.19#

#= color(blue)(10.81)#

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