Question

At 85°c, the ph of pure water is 6.25. calculate kw at this temperature.

Answer 1

Answer:
`3.2 * 10^(-13)`

Step-by-step explanation:

`K_w=[\text{H}^(+)] * [\text{OH}^(-)]`

However, as this is pure water,
`[\text{H}^(+)]=[\text{OH}^(-)]`, meaning that
`K_w=([\text{H}^(+)])^2`.

We can find
`[\text{H}^(+)]` from the pH.

`\text{pH}=-\log([\text{H}^(+)])\\\\6.25=-\log([\text{H}^(+)])\\\\`

`[\text{H}^(+)]=10^(-6.25)`.

Therefore,
`K_w=(10^(-6.25))^2 =3.2 * 10^(-13)`.