Question

What is the pH of pure water at 40.0°C if the Kw at this temperature is 2.92 A) 6.767 10-14? OB) 0.465 C)7.000 OD) 7.233 E) 8.446

Answer 1

Step-by-step explanation:

As we are given pure water. This means that concentration of hydrogen and hydroxide ions will be the same as pure water is neutral in nature.

Also, value of
`K_(w)` =
`2.92 * 10^(-14)`.

Relation between
`K_(w)` and concentration of hydrogen and hydroxide ions is as follows.

`K_(w) = [H^(+)][OH^(-)]`

Since,
`[H^(+)]` =
`[OH^(-)]`.

So,
`K_(w) = [H^(+)][OH^(-)]` =
`[H^(+)]^(2)`

`[H^(+)]` =
`\sqrt{K_(w)}`

=
`\sqrt{2.92 * 10^(-14)}`

=
`1.709 * 10^(-7)`

As we know that pH =
`-log[H^(+)]`

pH =
`-log(1.709 * 10^(-7))`

= 6.676

Thus, we can conclude that pH of pure water under given conditions is 6.676.