Calculate the pH of the solution of #10^(-6)# M #NaOH# at #50^o#C?

I got #"pH" = 7.28# for the exact solution, and #7.26# using the small #x# approximation.

This is still basic #"pH"#, as it is larger than #6.63#, neutral #"pH"# at #50^@ "C"#.

This has two complications:

• This is at #50^@ "C"#, so #K_w# is NOT #10^(-14)#.
• The concentration of #"NaOH"# put into water is not large, so the #"OH"^(-)# in water interferes.

At #50^@ "C"#, #K_w = 5.476 xx 10^(-14)#, so the #"pK"_w# is #13.26# and

#"pH" + "pOH" = 13.26#

Neutral #"pH"# would mean:

#["H"^(+)] = ["OH"^(-)] = sqrt(K_w) = 2.34 xx 10^(-7)#

And this neutral #"pH"# is

#"pH"_"neutral" = -log(2.34 xx 10^(-7)) = 6.63#

which is NOT #7# because we're warmer than #25^@ "C"#.

Also, we can see that #2.34 xx 10^(-7) "M OH"^(-)# is not much smaller than #10^(-6) "M"# #"OH"^(-)#. Therefore, we have a common ion effect.

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

#"I"" "-" "" "" ""0 M"" "" "" "" "" "10^(-6) "M"#
#"C"" "-" "" "+x" "" "" "" "" "" "+x#
#"E"" "-" "" "x" M"" "" "" "" "10^(-6) "M" + x#

The mass action expression then becomes:

#5.476 xx 10^(-14) = x(10^(-6) + x)#

We could solve this in full, but we can check the small #x# approximation.

#x^2 + 10^(-6)x - 5.476 xx 10^(-14) = 0#

The exact physical solution is #x = 5.205 xx 10^(-8) "M"#, so

#["H"^(+)] = 5.205 xx 10^(-8) "M"#

#color(blue)("pH") = -log["H"^(+)] = color(blue)(7.28)#

If #x# is small, then #["H"^(+)] = 5.476 xx 10^(-8) "M"# and #"pH" = 7.26#. So yes, the approximation is good.