Which is the pOH of an aqueous solution in which [OH–] = 0.00085M?Multiple choice question.3.003.07–3.00–3.07

1. Break Down the Problem

To find the pOH of the solution, we will use the relationship between hydroxide ion concentration $[OH^-]$ and pOH.

2. Relevant Concepts

The formula to calculate pOH is given by: $\text{pOH} = -\log[OH^-]$ Where $[OH^-]$ is the concentration of hydroxide ions in moles per liter (M).

3. Analysis and Detail

Given:

• $[OH^-] = 0.00085 \, M$

Calculating pOH: $\text{pOH} = -\log(0.00085)$ Using a calculator:

• Calculate $\log(0.00085)$: $\log(0.00085) \approx -3.069$ Thus, $\text{pOH} = -(-3.069) \approx 3.069$ Rounding to two decimal places, we have: $\text{pOH} \approx 3.07$

4. Verify and Summarize

The calculated pOH is approximately 3.07, confirming it is consistent with the logarithmic relationship for hydroxide ions.

Final Answer

The pOH of the aqueous solution is approximately 3.07.