Question

a 6.0 m base produces 0.25 m oh⁻ in solution. what is the percent ionization of oh⁻ from this base? 24% 42% 4.2% 0.24%

Explanation:

Step1: Recall percent - ionization formula

Percent ionization = $\frac{[\text{OH}^-]_{\text{produced}}}{[\text{Base}]_{\text{initial}}}\times100\%$

Step2: Substitute given values

Given $[\text{OH}^-]_{\text{produced}} = 0.25\ M$ and $[\text{Base}]_{\text{initial}}=6.0\ M$.
Percent ionization = $\frac{0.25}{6.0}\times 100\%$
Percent ionization = $\frac{25}{6.0}\% \approx 4.2\%$

Answer:

C. 4.2%