Buffer pH Change Trick in 60 Seconds! 💡 | NH₃–NH₄Cl Super Shortcut

 

❓ Question

A basic buffer contains:

$${\text{NH}}_3=0.10\text{ mol},{\text{NH}}_4^{+}=0.10\text{ mol}$$

Strong acid added:

$$\text{HCl}=0.05\text{ mol}$$

Given:

$$p{K}_b({\text{NH}}_3)=4.745$$

Find the change in pH after adding HCl.

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✍️ Short Solution

This is a weak base buffer (NH₃–NH₄Cl).
When HCl is added, it reacts completely with NH₃, converting it into NH₄⁺.
We then apply the Henderson–Hasselbalch equation for basic buffers to find the new pH.

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🔹 Step 1 — Buffer System

$$\text{NH}_3 + \text{NH}_4^+$$

Both initially = 0.10 mol
→ Perfect weak base buffer.

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🔹 Step 2 — Henderson–Hasselbalch Equation (Basic Buffer)

$$\text{pOH} = pK_b + \log\left(\frac{\text{salt}}{\text{base}}\right)$$$$\text{pH} = 14 - \text{pOH}$$

Given:

$$p{K}_b=4.745$$

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🔹 Step 3 — Reaction With Strong Acid

HCl reacts fully with NH₃:

$$\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4^+$$

HCl added = 0.05 mol

New moles:

$$\text{NH}_3 = 0.10 - 0.05 = 0.05\ \text{mol}$$$${\text{NH}}_4^{+}=0.10+0.05=0.15\text{ mol}$$

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🔹 Step 4 — New pH Using Henderson Equation

$$\text{pOH} = 4.745 + \log\left(\frac{0.15}{0.05}\right)$$$$\log 3 = 0.477$$

So,

$$\text{pOH} = 4.745 + 0.477 = 5.222$$
$$\text{pH}=14-5.222=8.778$$

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🔹 Step 5 — pH Change

Initial pH:

$$\text{pOH}_{\text{initial}} = 4.745 + \log\left(\frac{0.10}{0.10}\right) = 4.745$$
$$\text{pH}_{\text{initial}} = 14 - 4.745 = 9.255$$

Final pH = 8.778

Change in pH:

$$\Delta \text{pH} = 9.255 - 8.778 = 0.477$$
$$\Delta \text{pH} = 4.77 \times 10^{-2}$$

Nearest integer form:

$$\boxed{{\displaystyle 5\times {10}^{-2}}}$$

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✅ Final Answer

$$\boxed{{\displaystyle 5\times {10}^{-\mathrm{2}}}}$$

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