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
It does not matter what base you are looking at. If you know its
#"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 = x^2/(0.34 - x)#
In fact, the
#K_b ~~ x^2/0.34#
and the equilibrium concentration of
#x = sqrt(0.34K_b)#
#= 6.39 xx 10^(-4)# #"M"#
Therefore, the
#color(blue)("pH") = 14 - "pOH"#
#= 14 - (-log["OH"^(-)])#
#= 14 - 3.19#
#= color(blue)(10.81)#