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A 0.2 L of 0.05 M solution of dibasic (Na2HPO4) phosphate buffer is mixed with 0.16 L of 0.05 M monobasic (NaH2PO4) phosphate buffer. Calculate pH of the buffer solution

User Karptonite
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Final answer:

The pH of the buffer solution prepared from mixing 0.2 L of 0.05 M Na2HPO4 and 0.16 L of 0.05 M NaH2PO4 is 7.2, as calculated using the Henderson-Hasselbalch equation with the given moles of each species and pKa value.

Step-by-step explanation:

To calculate the pH of a buffer solution prepared from mixing a 0.2 L solution of 0.05 M Na2HPO4 (dibasic sodium phosphate) with a 0.16 L solution of 0.05 M NaH2PO4 (monobasic sodium phosphate), we use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Firstly, we need to calculate the moles of each species. For Na2HPO4: 0.2 L × 0.05 M = 0.01 mol, and for NaH2PO4: 0.16 L × 0.05 M = 0.008 mol. The pKa for the NaH2PO4/Na2HPO4 equilibrium is approximately 7.2.

Next, we use the Henderson-Hasselbalch equation to calculate the pH:

pH = 7.2 + log([0.01 mol / 0.2 L]/[0.008 mol / 0.16 L])

pH = 7.2 + log(0.05/0.05)

pH = 7.2

The pH of the buffer solution is 7.2.

User Gunaseelan
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