pecive cmouS 0 0otn dcia and comugate base present at each potnt amring. Dotr Ca
ID: 543852 • 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 acid = MV = 3*16 = 18
mmol of base = MV = 2*13 = 26
H3PO4 + OH- = H2PO4-
mmol of H3PO4 left = 18-18
mmol of OH- left = 26-18 = 8
mmol of H2PO4- formed = 26
H2PO4- + OH- --> HPO4-
mmol of OH- left = 8-8 = 0
mmol of H2po4- left = 26-8 = 16
mmol of HPO4-2 formed = 8
pH = pKa + log(HPO4-2/HPO3-)
pH = 7.21 + log(8/16)
ph = 6.91
b)
mol of HPO4-2 = mass/MW = 5/142 = 0.03521
mol of H+ = MV = 4*6*10^-3 = 0.024
after reaction
mol of H2PO4- = 0.024
mol o fHPO4-2 = 0.03521-0.024 = 0.01121
pH = pKa + log(HPO4-2/HPO3-)
pH = 7.21 + log(0.01121/0.024)
pH = 6.87939
c)
mol o fH2PO4 = mass/MW = 3.60/120 = 0.03
mol of HPO4-2 = mass/MW = 7.10/142 = 0.05
pH = pKa + log(HPO4-2/HPO3-)
pH = 7.21 + log(0.05/0.03)
pH = 7.4318
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