Question

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

Answer 1

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"`.

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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.