Question

What is the pH of \(2 \times 10^{-3}\ \text{M}\) solution of a monacidic weak base if it ionises to the extent of \(5\%\)?

• A. \(14\)
• B. \(10\)
• C. \(4\)
• D. \(2\)

Hint:

For weak bases: \[ [OH^-] = C\alpha \] where:
• \(C\) = concentration
• \(\alpha\) = degree of ionization Then use: \[ pOH = -\log[OH^-] \] and finally: \[ pH = 14 - pOH \] Always convert percentage ionization into decimal form before substitution.

Answer 1

The Correct Option is B

Concept: A weak base ionizes partially in aqueous solution to produce hydroxide ions \((OH^-)\). For a monacidic weak base: \[ BOH \rightleftharpoons B^+ + OH^- \] If:
• Initial concentration \(= C\)
• Degree of ionization \(= \alpha\) then concentration of hydroxide ions produced is: \[ [OH^-] = C\alpha \] After calculating \([OH^-]\), we use: \[ pOH = -\log[OH^-] \] and \[ pH + pOH = 14 \]

Step 1: Writing the given data.}
Concentration of weak base: \[ C = 2 \times 10^{-3}\ \text{M} \] Degree of ionization: \[ 5\% = \frac{5}{100} = 0.05 \]

Step 2: Calculating hydroxide ion concentration.}
\[ [OH^-] = C\alpha \] Substituting values: \[ [OH^-] = (2 \times 10^{-3})(0.05) \] \[ [OH^-] = (2 \times 10^{-3})\left(\frac{5}{100}\right) \] \[ [OH^-] = \frac{10 \times 10^{-3}}{100} \] \[ [OH^-] = 10^{-4}\ \text{M} \]

Step 3: Calculating pOH.}
\[ pOH = -\log(10^{-4}) \] \[ pOH = 4 \]

Step 4: Calculating pH.}
Using: \[ pH + pOH = 14 \] \[ pH + 4 = 14 \] \[ pH = 10 \]

Step 5: Final conclusion.}
Therefore, the pH of the solution is: \[ \boxed{10} \] Hence, the correct option is: \[ \boxed{(2)\ 10} \]