Problem 38

# Problem 38 - Problem 38(Note See problem1-36 in Segel Book...

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Problem 38 pH = pK a2 + log [HPO 4 = ] [H 2 PO 4 - ] Since the buffer will be at pH 6.9, the Bronsted base will be HPO 4 = and the Bronsted acid will be H 2 PO 4 - . Use the Henderson Hasselbalch equation as follows to determine the ratio of [Bronsted base] to [Bronsted acid].: 6.9 = 7.21 + log [HPO 4 = ] [H 2 PO 4 - ] -.31 = log [HPO 4 = ] [H 2 PO 4 - ] If x = 2 x = (Note: See problem1-36 in Segel Book, Page 57 for a similar problem.) or .31 = - log [H 2 PO 4 - ] [HPO 4 = ] (Note: I simply rearranged the equation by multiplying each side by -1 to make it easier to deal with and avoid getting a negative antilog.) 2 = log [H 2 PO 4 - ] [HPO 4 = ] [HPO 4 = ] [H 2 PO 4 - ] 40 liters x .02 moles/liter = .8 moles of phosphate is the total amount needed. To prepare 40 liters of .02 M phosphate buffer, you will need:: You will need .8 moles of base to convert H 3 PO 4 to H 2 PO 4 You will then have to add additional base to convert some of the H 2 PO 4 - to HPO 4 = To calculate this: x + 2x = .8 x = .2668 mole = amount of base to add 2x = .533 mole = amount of acid left

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Problem 38 - Problem 38(Note See problem1-36 in Segel Book...

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