KCET · Chemistry · Ionic Equilibrium
A \(0.15\) mole of pyridinium chloride has been added to \(500\text{ cm}^3\) of \(0.2\) M pyridine solution (a base). Assuming there is no change in volume upon mixing(Given: \(pK_b\) of pyridine = 8.82), the pH of the resulting solution is
- A \(5\)
- B \(6\)
- C \(7\)
- D \(8\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
Moles of pyridine (base) \(= 0.2 \times 0.5 = 0.1\) mol
Moles of pyridinium chloride (salt) \(= 0.15\) mol
Using the Henderson-Hasselbalch equation for a basic buffer:
\(pOH = pK_b + \log\left(\dfrac{\text{Moles of salt}}{\text{Moles of base}}\right)\)
\(pOH = 8.82 + \log\left(\dfrac{0.15}{0.10}\right)\)
\(pOH = 8.82 + \log(1.5)\)
\(pOH = 8.82 + 0.176 \approx 9\)
\(pH = 14 - pOH = 14 - 9 = 5\)
Answer: \(5\)
Moles of pyridinium chloride (salt) \(= 0.15\) mol
Using the Henderson-Hasselbalch equation for a basic buffer:
\(pOH = pK_b + \log\left(\dfrac{\text{Moles of salt}}{\text{Moles of base}}\right)\)
\(pOH = 8.82 + \log\left(\dfrac{0.15}{0.10}\right)\)
\(pOH = 8.82 + \log(1.5)\)
\(pOH = 8.82 + 0.176 \approx 9\)
\(pH = 14 - pOH = 14 - 9 = 5\)
Answer: \(5\)
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