1. In the laboratory, two forms of sodium phosphate will be available (the monob

Question

1. In the laboratory, two forms of sodium phosphate will be available (the monobasic monohydrate NaH2PO4*H2O, F.W = 137.99 g/mol, and the dibasic Na2HPO4, F.W = 141.96 g/mol). Explain the most efficient way of making this phosphate buffer by using only one of the compounds, and specify which compound you will use. 2. Provide calculations and explain how the 0.3% Al2(SO4)3 (w/v) solution will be prepared. 3. Provide calculations and explain how the 0.5M NaCl in 25mM sodium phosphate buffer, pH 8, will be prepared. 4. How much sodium phosphate will be required to make the 25mM sodium phosphate solution?

Explanation / Answer

The two forms of sodium phosphate are

NaH2PO4 and Na2HPO4

We will use NaH2PO4 as salt because if we will use NaH2PO4 and treate it with a base like NaOH the reaction will be

NaH2PO4 + NaOH --> Na2HPO4 + H2O

So if we will take moles of NaOH[1moles] less than the moles of NaH2PO4 [2 mole] it will give one mole of Na2HPO4 and one mole of NaH2PO4 will remain unreacted

So pH = pKa + log [salt] / [acid] = pKa + log [Na2HPO4] / [NaH2PO4 ]

So will get a buffer solution.

2) 0.3% of Al2(SO4)3 will be formed by dissolving 0.3 grams of Al2(SO4)3 in 100 of waterthe

4) for 25mM of sodium phosphate solution, we will dissolve 25millimoles of sodium phosphate in 1 L of water

molecular weight of sodium phosphate = 137.99 g /mole

So for 25millimoles the amount of sodium phosphate required will be

mass = 25 X 10^-3 X 137.99 grams = 3.45 grams will be dissolved in 100mL of solution

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