What is the pH of a 0.026 M Sr(OH)2 solution?

Well, what's the #"pOH"#? I get #"pOH" = 1.28#. What then is the #"pH"# at #25^@ "C"#?

We assume #"Sr"("OH")_2# is a strong base, such that

#"Sr"("OH")_2(s) stackrel("H"_2"O"(l)" ")(->) "Sr"^(2+)(aq) + 2"OH"^(-)(aq)#

and thus, it supposedly gives rise to #0.026 xx 2 = "0.052 M OH"^(-)#. As a result,

#"pOH" = -log["OH"^(-)] = -log(0.052) = 1.28#

But clearly, we have the #"pOH"# and not the #"pH"#. At any temperature,

#"pH" + "pOH" = "pK"_w#,

and at #25^@ "C"#, #"pK"_w = 14#. Therefore:

#color(blue)("pH") = 14 - 1.28 = color(blue)(12.72)#

But as chemists, we must check the data... The #K_(sp)# of #"Sr"("OH")_2# is around #6.4 xx 10^(-3)#. The ICE table gives:

#"Sr"("OH")_2(s) rightleftharpoons "Sr"^(2+)(aq) + 2"OH"^(-)(aq)#

#"I"" "-" "" "" "" "" "0" "" "" "" "0#
#"C"" "-" "" "" "" "+s" "" "" "+2s#
#"E"" "-" "" "" "" "" "s" "" "" "" "2s#

This is equal to the mass action expression:

#K_(sp) = 6.4 xx 10^(-3) = ["Sr"^(2+)]["OH"^(-)]^2#

#= s(2s)^2 = 4s^3#

#= 1/2(2s)^3 = 1/2["OH"^(-)]^3#

And so, the maximum concentration of #"OH"^(-)# at #25^@ "C"# is:

#["OH"^(-)] = (2K_(sp))^(1//3)#

#= (2 cdot 6.4 xx 10^(-3))^(1//3)#

#=# #"0.234 M"#

And since #"0.052 M"# #<# #"0.234 M"#, the concentration given in the question is valid and physically realizable.