Question

What is the pOH of a solution at 25.0∘C with [H3O+]=4.8×10−6 M?

Answer 1

8.68

Step-by-step explanation:

pOH = 8.68

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Answer 2

Approximately
`8.68`.

Step-by-step explanation:

The
`\rm pOH` of a solution can be found from the hydroxide ion concentration
`\rm \left[OH^(-)\right]` with the following equation:

`\displaystyle \rm pOH = -\log_(10) \rm \left[OH^(-)\right]`.

On the other hand, the ion-product constant of water,
`K_{\text{w}}`, relates the hydroxide ion concentration
`\rm \left[OH^(-)\right]` of a solution to its hydronium ion concentration
`\rm \left[{H_3O}^(+)\right]`:

`K_\text{w} = \rm \left[{H_3O}^(+)\right] \cdot \rm \left[OH^(-)\right]`.

• At
  `25 \; ^\circ \rm C`,
  `K_{\text{w}} \approx 1.0 * 10^(-14)`.
• For this particular
  `25 \; ^\circ \rm C` solution,
  `\rm \left[{H_3O}^(+)\right] = 4.8 * 10^(-6)\; \rm mol \cdot L^(-1)`.

Hence the
`\rm \left[OH^(-)\right]` of this solution:

`\begin{aligned}\left[\mathrm{OH}^(-)\right] &= \frac{K_\text{w}}{\rm \left[{H_3O}^(+)\right]} \\ &= (1.0 * 10^(-14))/(4.8 * 10^(-6))\; \rm mol\cdot L^(-1) \approx 2.08333 * 10^(-9)\; \rm mol\cdot L^(-1)\end{aligned}`.

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

`\begin{aligned}\rm pOH &= -\log_(10) \rm \left[OH^(-)\right] \\ &\approx -\log_(10) \left(4.8 * 10^(-6)\right) \approx 8.68\end{aligned}`.

Note that by convention, the number of decimal places in
`\rm pOH` should be the same as the number of significant figures in
`\rm \left[OH^(-)\right]`.

For example, because the
`\rm \left[{H_3O}^(+)\right]` from the question has two significant figures, the
`\rm \left[OH^(-)\right]` here also has two significant figures. As a result, the
`\rm pOH` in the result should have two decimal places.