Question

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

Answer 1

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

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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`...