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newlabguy
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This isn't homework. It's for my job. I'm supposed to be making a 50mM phosphate buffer to pH 7.2, 0.5 NaOH. I have access to salts, K2HPO4 (dibasic), KH2PO4 (monobasic), and NaCl. I've determined that I need two liters of buffer solution. I did the calculations but when I measured the pH after making the solution, it was more acidic than it should be. I adjusted the pH with strong base, so it's alright now. I'm just wondering if someone can verify my calculations.
FW K2HPO4 = 174.18 g/mol
FW KH2PO4 = 136.09 g/mol
FW NaCl = 58.44 g/mol
pH = pKa + log ([K2HPO4]/[KH2PO4])
Since pKa = 7.2 and pH =7.2, the HH equation reduces to [KH2PO4]=[K2HPO4]
For two liters, I need 0.1 mol phosphate buffer
0.1 mol= [KH2PO4]+[K2HPO4]
then
0.05 mol = [KH2PO4]
0.05 mol = [K2HPO4]
then I weigh out 136.09 g/mol * 0.05 mol = 6.8045 g [KH2PO4]
and 174.18 g/mol * 0.05 mol = 8.709 g [K2HPO4]
and 58.44 g NaCl
and fill the volume to 95% and adjust pH if necessary then fill to mark. Did I do something wrong? Some buffer calculators are showing something like 3 times the amount of base as acid and I don't know why. For example, http://www.biomol.net/en/tools/buffercalculator.htm
FW K2HPO4 = 174.18 g/mol
FW KH2PO4 = 136.09 g/mol
FW NaCl = 58.44 g/mol
pH = pKa + log ([K2HPO4]/[KH2PO4])
Since pKa = 7.2 and pH =7.2, the HH equation reduces to [KH2PO4]=[K2HPO4]
For two liters, I need 0.1 mol phosphate buffer
0.1 mol= [KH2PO4]+[K2HPO4]
then
0.05 mol = [KH2PO4]
0.05 mol = [K2HPO4]
then I weigh out 136.09 g/mol * 0.05 mol = 6.8045 g [KH2PO4]
and 174.18 g/mol * 0.05 mol = 8.709 g [K2HPO4]
and 58.44 g NaCl
and fill the volume to 95% and adjust pH if necessary then fill to mark. Did I do something wrong? Some buffer calculators are showing something like 3 times the amount of base as acid and I don't know why. For example, http://www.biomol.net/en/tools/buffercalculator.htm
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