pecive cmouS 0 0otn dcia and comugate base present at each potnt amring. Dotr Ca
ID: 543853 • Letter: P
Question
pecive cmouS 0 0otn dcia and comugate base present at each potnt amring. Dotr Carcutations 5. There are many ways for making a phosphate buffer. Depending on what is available in the lab, (ii) a phosphate salt and the addition of either a strong acid or base, eg. NaH2PO4/ NaOH; (ii) the phosphate salt of both the conjugate acid and the conjugate base, eg. NaH2PO4 /NazHPO4. or (iv) or a combination of the reagents mentioned above. Most of the time we start with a solution of the salt form and then titrate with either a strong base or acid to obtain the pH that we desire. Determine the pH and the final concentration of the following buffer solutions (remember to show your calculation and comment on your rational). (a) 6.00 mL of 3.0 M HsPO, 13.0 ml of 2.0 M NaOH and enough water to make a final volume of 100 ml. (b) 5.00 g of Na,HPO (MW 142 g/mol), 4.00 ml of 6.0 M HCI and enough water to make a final volume of 200 ml. (c) 3.60 g of NaH PO, (MW 120 g/mol) and 210 g Na,HPO, (MW 142 g/mol) dissolved in a total volume of 0.400 litre of water. Buffers-13Explanation / Answer
a)
mmol of H3PO4 = MV = 6*3 = 18
mmol of base = 13*2 = 26
clearly
H3PO4+ OH- = H2PO4-
mmol of H3PO4 left = 0
mmol of NaOH = 26-18 = 8
after 2nd neutraliatoin
H2PO4- + OH --> HPO4-
H2PO4- left = 18-8 = 10
HPO4-2 produced = 8
pH = pKa2 + log(HPO4-2/H2PO4)
pH = 7.21 + log(8/10)
pH = 7.1130
b)
mol of HPO4- = mass/MW = 5/142 = 0.0352
mol of HCl = MV = (4*10^-3)(6) = 0.0240
mol of HPO4- left = 0.0352-0.0240 = 0.0112
mol o fH2PO4- produced = 0.0240
substitute in pH
pH = pKa2 + log(HPO4-2/H2PO4)
pH = 7.21 + log(0.0112/0.0240) =
pH = 6.88
c)
mol of NaH2PO4 = mass/MW = 3.60/120 = 0.03
mol of Na2HPO4 = mass/MW = 7.10/142 = 0.05
this is a buffer so
pH = pKa2 + log(HPO4-2/H2PO4)
pH = 7.21 + log(0.05/0.03)
pH = 7.431
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