millimoles acetic acid = 1000 mL x 0.01M = 10
mmole sodium acetate = 1000 mL x 0.1M = 100.
5.00 = 4.74 + log(base/acid)
base = 100-x
acid = 10+x
Solve for x where x = mols HNO3 added.
Then convert to volume needed of the 10M HNO3.
how much 10M HNO3 must be added to 1L of a buffer that is .010M acetic acid and .10M sodium acetate to reduce the pH to 5. pKa=4.75
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