What is the pH of #1.0 xx 10^(-10)# #"M HCl"#?

1 Answer
Truong-Son N.
Apr 7, 2018

Well, it's not #10#...

The concentration is so small that water should overwhelm it, but let's see.

#2"H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "OH"^(-)(aq)#

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

Now water autoionizes to give at #25^@ "C"#

#K_w = 10^(-14) = ["H"_3"O"^(+)]["OH"^(-)]#

#= (10^(-10) + x)x#

Here #x# will not be small compared to #10^(-10)#, but #10^(-10) "M"# may be small compared to #x#.

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

The exact solution is #x = 9.995 xx 10^(-8) "M"#. The approximation that #x ">>" 10^(-10) "M"# would give #x = 10^(-7) "M"#, which isn't that far off. Using the exact solution,

#["H"_3"O"^(+)] = 1.0000 xx 10^(-10) "M" + 9.9950 xx 10^(-8) "M"#

#= 1.0005 xx 10^(-7) "M"#

And this would give #color(blue)("pH" = 6.9998)#, which is very slightly acidic to four decimal places, but you might as well call it #7#...

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