Question

The $\mathrm{pH}$ of monoacidic weak base is $10.9$. Calculate the percent dissociation in $0.02\ \mathrm{M}$ solution.

• A. $7.92\%$
• B. $3.95\%$
• C. $6.25\%$
• D. $2.51\%$

Hint:

When computing $\text{antilog}(-3.1)$, split it into integers and decimals: $10^{-3.1} = 10^{-4} \times 10^{0.9}$. Since $\log_{10}(8) \approx 0.903$, $10^{0.9}$ is slightly less than $8$ ($\approx 7.94$). This approximation saves immense time during exams.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the $\mathrm{pH}$ ($10.9$) of a monoacidic weak base solution with a concentration ($c$) of $0.02\ \mathrm{M}$. We need to determine its percent dissociation.

Step 2: Key Formula or Approach:
First, find $\mathrm{pOH}$ using the relationship: $$\mathrm{pH} + \mathrm{pOH} = 14 \implies \mathrm{pOH} = 14 - \mathrm{pH}$$ Next, compute the hydroxide ion concentration: $$[\mathrm{OH^-}] = 10^{-\mathrm{pOH}}$$ For a monoacidic weak base, the relationship between $[\mathrm{OH^-}]$, degree of dissociation ($\alpha$), and concentration ($c$) is: $$[\mathrm{OH^-}] = \alpha \cdot c \implies \alpha = \frac{[\mathrm{OH^-}]}{c}$$ Finally, evaluate percent dissociation using $\% \alpha = \alpha \times 100\%$.

Step 3: Detailed Explanation:
Calculate $\mathrm{pOH}$: $$\mathrm{pOH} = 14 - 10.9 = 3.1$$ Calculate $[\mathrm{OH^-}]$: $$[\mathrm{OH^-}] = 10^{-3.1} = \text{antilog}(-3.1) = \text{antilog}(\bar{4}.9) \approx 7.943 \times 10^{-4}\ \mathrm{M}$$ Now, find the degree of dissociation $\alpha$: $$\alpha = \frac{7.943 \times 10^{-4}}{0.02} = 3.97 \times 10^{-2}$$ Convert to a percentage value: $$\% \alpha = 3.97 \times 10^{-2} \times 100 = 3.97\%$$ This value is closest to $3.95\%$.

Step 4: Final Answer:
The percent dissociation of the monoacidic weak base is $3.95\%$, which matches option (B).