please answer based on the chart provided. Thanks 5.0 Electrolyte Preparation Th
ID: 498126 • Letter: P
Question
please answer based on the chart provided. Thanks
Explanation / Answer
1) Given that
[NaH2PO4] = 0.5 mM
[Na2HPO4] = 2 mM
We know that pKa of NaH2PO4 = 7.2
From Henderson-hasselbalch equation,
pH = pKa + log [conjugate base] /[acid]
= pKa + log [Na2HPO4] /[NaH2PO4 ]
= 7.2 + log ( 2 mM/ 0.5 mM)
= 7.8
pH = 7.8
Therefore,
pH of the buffer = 7.8
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