Calculate the pH of a saturated aqueous solution of #Co(OH)_3#(#K_(sp) = 2.7*10^(-43)#)?

1 Answer
Truong-Son N.
Mar 7, 2018

Well, it ought to be close to #7#. That's quite a small #K_(sp)#... I get #7.000065# or so. Still basic, but who really cares about the fifth decimal place of #"pH"#?

Write the reaction:

#"Co"("OH")_3(s) rightleftharpoons "Co"^(3+)(aq) + 3"OH"^(-)(aq)#

This gives a #K_(sp)# expression of:

#2.7 xx 10^(-43) = ["Co"^(3+)]["OH"^(-)]^3#

#= s(3s)^3 = 27s^4#

As a result,

#["OH"^(-)] = 3s = 3 cdot ((2.7 xx 10^(-43))/27)^(1//4)#

#= 3.0 xx 10^(-11) "M"#

This is negligible compared to the #"OH"^(-)# found in water at #25^@ "C"#, so #color(blue)("pH" = 7)#. Here's the exact solution...

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

#"I"" "-" "" "" "0" "" "" "" "3.0 xx 10^(-11)#
#"C"" "-" "" "+x" "" "" "+x#
#"E"" "-" "" "x" "" "" "" "3.0 xx 10^(-11) + x#

Thus:

#K_w = 10^(-14) = x(3.0 xx 10^(-11) + x)#

#=> x^2 + 3.0 xx 10^(-11)x - 10^(-14) = 0#

For this, #x = 9.99850011 xx 10^(-8) "M"#. As a result,

#["H"^(+)] = 9.99850011 xx 10^(-8) "M"#

#=> "pH" = -log["H"^(+)] = 7.000065 ~~ color(blue)(7)#

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