1. what is the poh of a solution that has a hydroxide ion concentration of 5.64 x 10-6 m? express your…

Explanation:

Step1: Substitute the given value.

Given $[\text{OH}^-]=5.64\times 10^{-6}\text{ M}$, and the formula $p\text{OH}=-\log[\text{OH}^-]$. So we have $p\text{OH}=-\log(5.64\times 10^{-6})$.

Step2: Use logarithm properties.

Using the property $\log(ab)=\log a+\log b$, we get $p\text{OH}=-(\log(5.64)+\log(10^{-6}))$. Since $\log(10^{-6})=- 6$, then $p\text{OH}=-( \log(5.64)-6)$. And $\log(5.64)\approx0.751$. So $p\text{OH}=-(0.751 - 6)$.

Step3: Calculate the result.

$p\text{OH}=6 - 0.751=5.249\approx5.25$.

Answer:

$5.25$