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Item 1: Question (10 pts.) What is the pH of the buffer that results when 13.4 g of NaH,PO, and 5.09 g of Na HPO4 are diluted
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Answer #1

5.09 - M 142.080.912 13.4 [Naha Pon] = m [Na₂HPO4] =- 120.0 40.912 Using HH equation, PH = pka + log [Nag HP On] IN GHz PDy]

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