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In a basic buffer, 0.0025 mole of 'NH4 ...

In a basic buffer, 0.0025 mole of 'NH_4 Cl' and 0.15 mole of `NH_4OH` are present. The pH of the solution will be `(pK_a)=4.74`

A

11.04

B

10.24

C

6.62

D

5.48

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the pH of a basic buffer solution containing ammonium chloride (NH4Cl) and ammonium hydroxide (NH4OH), we can use the Henderson-Hasselbalch equation. Here’s a step-by-step solution: ### Step 1: Identify the components of the buffer solution In the given problem, we have: - Salt: Ammonium chloride (NH4Cl) - Base: Ammonium hydroxide (NH4OH) ### Step 2: Write the Henderson-Hasselbalch equation The Henderson-Hasselbalch equation for a basic buffer is given by: \[ \text{pOH} = \text{pK}_B + \log \left( \frac{[\text{Salt}]}{[\text{Base}]} \right) \] ### Step 3: Determine the concentrations of salt and base We need to calculate the concentrations of the salt and base. - Moles of NH4Cl = 0.0025 moles - Moles of NH4OH = 0.15 moles Let the volume of the solution be \( V \) liters. Thus, the concentrations are: - Concentration of NH4Cl (Salt) = \( \frac{0.0025 \text{ moles}}{V \text{ L}} \) - Concentration of NH4OH (Base) = \( \frac{0.15 \text{ moles}}{V \text{ L}} \) ### Step 4: Substitute the concentrations into the Henderson-Hasselbalch equation Now we substitute the concentrations into the equation: \[ \text{pOH} = \text{pK}_B + \log \left( \frac{0.0025/V}{0.15/V} \right) \] This simplifies to: \[ \text{pOH} = \text{pK}_B + \log \left( \frac{0.0025}{0.15} \right) \] ### Step 5: Calculate the ratio Now, calculate the ratio: \[ \frac{0.0025}{0.15} = \frac{1}{60} \] ### Step 6: Calculate the logarithm Now we find the logarithm: \[ \log \left( \frac{1}{60} \right) = -\log(60) \approx -1.78 \] ### Step 7: Substitute the pK_B value and calculate pOH Given that \( \text{pK}_B = 4.74 \): \[ \text{pOH} = 4.74 - 1.78 = 2.96 \] ### Step 8: Calculate pH from pOH Finally, to find the pH: \[ \text{pH} = 14 - \text{pOH} = 14 - 2.96 = 11.04 \] ### Final Answer The pH of the solution is **11.04**. ---
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