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Problems · Problem 6.22

Q.Calculate the pH of the solution in which 0.2M NH4Cl and 0.1M NH3 are present. The pKb of ammonia solution is 4.75.

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This is a basic buffer solution (NH₃/NH₄⁺). Using the Henderson–Hasselbalch equation for a base buffer, pOH=pKb+log⁡[NH4+][NH3]\text{pOH} = \text{p}K_b + \log\frac{[\text{NH}_4^+]}{[\text{NH}_3]}, we get pOH = 4.75 + log(2) = 5.05, so pH = 14 – 5.05 = 8.95.

The mixture contains a weak base (NH₃) and its conjugate acid (NH₄⁺ from NH₄Cl). That is the classic composition of a basic buffer — a solution that resists changes in pH. The key idea is that the equilibrium concentration of NH₃ and NH₄⁺ are essentially the same as their initial concentrations because the common ion effect suppresses dissociation.

For any buffer, the pH is governed by the Henderson–Hasselbalch equation. Since we are dealing with a base and its conjugate acid, it is more direct to work with pOH first.

For a basic buffer:

pOH=pKb+log⁡[conjugate acid][base]\text{pOH} = \text{p}K_b + \log\frac{[\text{conjugate acid}]}{[\text{base}]}

Here, the base is NH₃ (0.1 M) and the conjugate acid is NH₄⁺ (0.2 M, from NH₄Cl). The pKb is given as 4.75.

  1. Identify the components

    NH₃ is the weak base. NH₄Cl is a salt that dissociates completely to give NH₄⁺ ions. So the solution has:

    • [NH3]=0.1 M[\text{NH}_3] = 0.1\ \text{M}
    • [NH4+]=0.2 M[\text{NH}_4^+] = 0.2\ \text{M}
  2. Apply the Henderson–Hasselbalch for pOH

pOH=4.75+log⁡0.20.1=4.75+log⁡2\text{pOH} = 4.75 + \log\frac{0.2}{0.1} = 4.75 + \log 2

Since log⁡2≈0.30\log 2 \approx 0.30,

pOH=4.75+0.30=5.05\text{pOH} = 4.75 + 0.30 = 5.05

  1. Convert to pH At 25°C, pH + pOH = 14. pH=14−5.05=8.95\text{pH} = 14 - 5.05 = 8.95 …

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