Question

what is the poh of a solution with oh⁻ = 2.0 × 10⁻²?
a. 3.01
b. 1.70
c. -1.70
d. 2.00

Explanation:

Step1: Recall pOH formula

The formula for pOH is $pOH = -\log[OH^-]$.

Step2: Substitute the given value

Given $[OH^-]=2.0\times 10^{-2}$, then $pOH = -\log(2.0\times 10^{-2})$.
Using the logarithm property $\log(ab)=\log a+\log b$, we have $pOH=-(\log 2.0+\log(10^{-2}))$.
Since $\log(10^{-2})=- 2$ and $\log 2.0\approx0.301$, then $pOH=- (0.301 - 2)$.

Step3: Calculate the result

$pOH=-0.301 + 2=1.70$.

Answer:

B. 1.70