Question

Determine the pH of a 7 × 10−5 M HNO3 solution. Your answer must be within ± 0.1%?

Answer 1

This concentration is large enough that

`"pH" = -log["H"_3"O"^(+)] = -log(7 xx 10^(-5)) = 4.15_(490196)`

This is clearly within `0.1%` of the exact solution:

`|(4.15490196 - 4.15490107)/(4.15490107)| xx 100% = 2.14 xx 10^(-5) %`

We can do it this way because in water, `["H"_3"O"^(+)] = 10^(-7) "M"` at `25^@ "C"`... and this is over `100` times bigger.

---

This could done using an ICE table for water... Why? Well, the concentration is very small.

`["HNO"_3] ~~ ["H"^(+)] = 7 xx 10^(-5) "M"`

This `"H"^(+)` interferes with the autoionization of water, which may or may not be significant.

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

`"I"" "-" "" "" "7 xx 10^(-5)" "" "" "0`
`"C"" "-" "" "+x" "" "" "" "" "+x`
`"E"" "-" "" "7 xx 10^(-5) + x" "" "x`

At `25^@ "C"`, `K_w = 10^(-14)`, so:

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

`= (7 xx 10^(-5)+x)x`

Compared to in water, `x` `"<<"` `7 xx 10^(-5)`, so we estimate that

`10^(-14) ~~ 7 xx 10^(-5)x`

`=> x = 1.43 xx 10^(-10) "M"`
(This is in fact, the exact solution to `x`.)

This is for `["OH"^(-)]`, but

`["H"^(+)] = 7 xx 10^(-5) + 1.43 xx 10^(-10) "M"`

Hence,

`color(blue)("pH" = 4.15)_(490107)`

where subscripts indicate digits past the reasonable digit of uncertainty.

This is as close as we could get to the exact solution.