based on log rules and the way ph is calculated, what is the difference in oh⁻ concentration between point a…
based on log rules and the way ph is calculated, what is the difference in oh⁻ concentration between point a and point b.\n10¹\n10⁵\n10⁶\n10⁷
Answer
Explanation:
Step1: Recall pOH - [OH⁻] relationship
The formula for pOH is $pOH = -\log[OH^{-}]$, so $[OH^{-}]=10^{-pOH}$.
Step2: Determine pOH values at A and B
At point A, $pH = 0$, so $pOH=14 - pH=14$. At point B, $pOH = 7$.
Step3: Calculate [OH⁻] at A and B
At point A, $[OH^{-}]_A = 10^{-14}$. At point B, $[OH^{-}]_B=10^{-7}$.
Step4: Find the ratio of [OH⁻] concentrations
$\frac{[OH^{-}]_B}{[OH^{-}]_A}=\frac{10^{-7}}{10^{-14}} = 10^{(- 7-(-14))}=10^{7}$.
Answer:
$10^{7}$