Question

What is the pH of a `2.6*10^-9` `M` `H^+` solution?

Answer 1

The pH is 6.99.

Explanation:

In pure water, `["H"^+] = ["OH"^"-"] = 1.00 × 10^"-7" color(white)(l)"mol/L"`.

Imagine that we add the `2.6 × 10^"-9" color(white)(l)"mol/L H"^+`, but the equilibrium hasn't had time to re-establish itself.

We have just set up a new initial condition.

Let's put this into an ICE table.

`color(white)(mmmmm)"H"_2"O" ⇌ color(white)(mmmmmll)"H"^+color(white)(mmmm) +color(white)(mll) "OH"^"-"`
`"I/mol·L"^"-1": color(white)(mmmm)1.00×10^"-7" + 2.6×10^"-9" color(white)(ml)1.00 × 10^"-7"`
`"C/mol·L"^"-":color(white)(mmmmmmmmm) -x color(white)(mmmmmmml)-x`
`"E/mol·L"^"-1":color(white)(mmmmm) 1.026 × 10^"-7" -xcolor(white)(mml) 1.00 × 10^"-7" - x`

`K_w = ["H"^+]["OH"^"-"] = 1.00 × 10^"-14"`

`(1.026 × 10^"-7" -x)(1.00 × 10^"-7" - x) = 1.00 × 10^"-14"`

`x^2 - 2.026 × 10^"-7"x + 1.026 × 10^"-14" = 1.00 × 10^"-14"`

`x^2 - 2.026 × 10^"-7"x + 0.026 × 10^"-14" = 0`

`x = 1.30 × 10^"-9"`

`["H"^+] = 1.026 × 10^"-7" -x = 1.026 × 10^"-7" - 1.03 × 10^"-9" = 1.016 × 10^"-7"`

`"pH" = -log["H"^+] = -log(1.016 × 10^"-7") = 6.99`