Question

If `[H_3O^+] = 1.005 * 10^-5` M, what is the pH?

Answer 1

`4.9978`

Explanation:

Right from the start, you can say that the pH of this solution will be very close to `5`.

Here's why that is so.

As you know, the pH of pure water at room temperature is equal to `7`. This value is derived from the self-ionization of water, in which two water molecules exchange a proton, `"H"^(+)`, to form hydronium and hydroxide ions

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

At room temperature, the ion product constant---which you may see as `K_w`---for this reaction is equal to `10^(-14)`, which is why

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

`["H"_3"O"^(+)] = ["OH"^(-)] = sqrt(10^(-14)) = 10^(-7)"M"`

As you know, pH is defined as

`"pH" = -log( ["H"_3"O"^(+)])`

For pure water, this is equal to

`"pH" = -log(10^(-7)) = 7`

In your case, assuming that you're at room temperature as well, the concentration of hydronium ions is higher than `10^(-7)`. This tells you that

• the solution is acidic**
• its pH must be lower than `7`

Since pH is defined as the negative log base `10` of the concentration of hydronium ions, a hydronium concentration that is approximately two times higher will result in a pH that is approximately two units lower.

`"pH" = -log(1.005 * 10^(-5)) = color(blue)(4.9978)`

Indeed, the resulting pH is very close to `5`, since `1.005 * 10^(-5)` is very close to `10^(-5)`.