Question

What is the pOH of a
2.6 x 10-6 M H+ solution?

Answer 1

Approximately
`8.41` (assuming that the solution is at
`\rm 25^\circ C`, under which
`K_(\rm w) = 10^(-14)`.)

Step-by-step explanation:

Let
`{\rm [H^(+)]}` and
`{\rm [OH^(-)]}` denote the concentration of
`\rm H^(+)` and
`\rm OH^(-)` respectively.

Let
`K_(\rm w)` denote the self-ionization constant of water. The exact value of
`K_(\rm w)\!` depends on the temperature of the solution.
`K_(\rm w) =10^(-14)` at
`\rm 25^\circ C`.

The product of
`{\rm [H^(+)]}` and
`{\rm [OH^(-)]}` in a solution (with
`\rm M`, or moles per liter, as the unit) is supposed to be equal to the
`K_(\rm w)` value of that solution at the corresponding temperature. In other words:

`{\rm [H^(+)]} \cdot {\rm [OH^(-)]} = K_(\rm w)`.

Rearrange to obtain an expression for
`{[\rm OH^(-)]}`:

`\begin{aligned}{\rm [OH^(-)]} &= (K_(\rm w))/([\rm H^(+)])\end{aligned}`.

Assume that the solution in this question is at
`\rm 25^\circ C` (for which
`K_(\rm w) =10^(-14)`.) For this solution:

`\begin{aligned}{\rm [OH^(-)]} &= (K_(\rm w))/([\rm H^(+)]) \\ &= (10^(-14))/(2.6 * 10^(-6))\approx 3.85* 10^(-9)\; \rm M\end{aligned}`.

Hence, the
`\rm pOH` of this solution would be:

`\begin{aligned}\rm pOH &= -\log_(10){\rm [OH^(-)]} \\&\approx -\log_(10) (3.85 * 10^(-9)) \approx 8.41 \end{aligned}`.