Q.41 One litre of phosphate buffer was prepared by adding 208 grams of Na2HPO4 (Mol. wt. 142) and
71 grams of NaH2PO4 (Mol. wt. 120) in water. If the pKa for the dissociation of H2PO4− into
HPO42− and H+ is 6.86, the pH of the buffer will be __________.
Given Data
- Volume of buffer = 1 L
- Mass of Na2HPO4 = 208 g
- Molecular weight of Na2HPO4 = 142
- Mass of NaH2PO4 = 71 g
- Molecular weight of NaH2PO4 = 120
- pKa = 6.86
Concept: Phosphate Buffer System
In a phosphate buffer:
- Acid (HA): H2PO4−
- Conjugate base (A−): HPO42−
The pH of the buffer is calculated using the
Henderson–Hasselbalch equation.
Step 1: Calculate Number of Moles
Moles of Na2HPO4 (Base)
Moles = 208 / 142 = 1.465 mol
Moles of NaH2PO4 (Acid)
Moles = 71 / 120 = 0.592 mol
Step 2: Apply Henderson–Hasselbalch Equation
pH = pKa + log ( [Base] / [Acid] )
pH = 6.86 + log (1.465 / 0.592)
pH = 6.86 + log (2.47)
log (2.47) ≈ 0.392
Final Answer
pH = 6.86 + 0.392 ≈ 7.25
Common Mistakes to Avoid
- Using mass instead of moles in calculations
- Reversing acid and base in the log term
- Using an incorrect pKa value
- Not applying the Henderson–Hasselbalch equation
Conclusion
By converting the given masses into moles and applying the
Henderson–Hasselbalch equation, the pH of the phosphate buffer
is calculated to be approximately 7.25.
This confirms that the buffer is slightly basic and suitable
for biological and biochemical applications.
