Question

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%

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%