a 6.0 m base produces 6.0 m oh⁻ in solution. what is the percent ionization of oh⁻ from this base? 60% 6% 1%…
a 6.0 m base produces 6.0 m oh⁻ in solution. what is the percent ionization of oh⁻ from this base? 60% 6% 1% 100%
Answer
Explanation:
Step1: Recall percent - ionization formula
Percent ionization of a base = $\frac{[\text{OH}^-]{\text{produced}}}{[\text{Base}]{\text{initial}}}\times100%$
Step2: Substitute given values
Given $[\text{Base}]{\text{initial}} = 6.0\ M$ and $[\text{OH}^-]{\text{produced}}=6.0\ M$. So, percent ionization = $\frac{6.0}{6.0}\times100%$
Step3: Calculate the result
$\frac{6.0}{6.0}\times100% = 100%$
Answer:
D. 100%