The pKa’s of the three acid-conjugate base pairs derived from phosphoric acid, H
ID: 1029093 • Letter: T
Question
The pKa’s of the three acid-conjugate base pairs derived from phosphoric acid, H3PO4, are 2.12, 7.21, and 12.32. How many g of sodium hydroxide would have t be added to 150mL of 0.1 M H3PO4 to prepare a buffer of pH = 7.41? The answer should be 0.83g but I can’t figure it how to get that. The pKa’s of the three acid-conjugate base pairs derived from phosphoric acid, H3PO4, are 2.12, 7.21, and 12.32. How many g of sodium hydroxide would have t be added to 150mL of 0.1 M H3PO4 to prepare a buffer of pH = 7.41? The answer should be 0.83g but I can’t figure it how to get that. The answer should be 0.83g but I can’t figure it how to get that.Explanation / Answer
The value of the pH = 7.41 is nearer to one of the second dissociation pKa2 = 7.21, which will give the required buffer. Therefore H3PO4 needs to be convereted first into the monodeprotonated form.
NaH2PO4 and H3PO4 combination buffer is required. They have already given hint in the question that "How many ----------------------------------------------------- H3PO4". We have to use H3PO4 acid as one of the combination so definitely its conjugate base will be NaH2PO4 only.
moles present in 150mL of 0.1 M H3PO4 acid = 150 mL x 10-3 L x 0.1M = 0.015 mol
from Henderson-Hasselbalch equation we have
pH = pKa + log[base / acid]
7.41 = 7.21 + log[base / 0.015]
log[base / 0.015] = 0.2
[base / 0.015] = 100.2
[base / 0.015] = 1.5849
base = 0.02377 moles required.
convert moles NaOH to mass NaOH by multiplying with molar mass of NaOH
mass of NaOH in g required = 39.997 g/mol x 0.02377 mol = 0.95 grams
I am sorry friend, standard protocol gave me answer as 0.95 g (But you have mentioned as 0.83 g, i do not know from where you got this value). Please don't try to look for a method to get required answer, it leads us to manipulation.
Hope this helped you!
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