# 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)#