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13 Dec 2019

What volumes of 0.5 M H3PO4 and 0.5 M NaOH are needed to make a 1.0 L solution buffered at a pH of 7.2. Ka for H3PO4 = 7.5 x 10^-3

I know I have to use the Henderson-Hasselbach Equation, but I'm not sure how to solve this problem. Thanks in advance!

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Jarrod Robel
Jarrod RobelLv2
17 Dec 2019

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Related questions


For question 3 part B how do we know when to add or subract to the acid or base in the Henderson hasselbach equation?

A liter of pH 7.20 phosphate buffer is needed for a certain experiment. The Henderson-Hasselbach equation will be sufficiently accurate for your determination of pH. The pK’s for possibly relevant phosphate species are:

H3PO4 ↔ H2PO4- + H+ pK = 2.15

H2PO4- ↔ HPO4-2 + H+ pK = 7.20

HPO4-2 ↔ PO4-3 + H+ pK = 12.4

0.100 Moles of H3PO4 were dissolved in about 800 mL of water, and the pH was adjusted to 7.20 using a standardized pH meter and a 1 M solution of KOH followed by addition of water to a volume of 1.00 L.

a. (10 pts) Write the equation for the charge balance to show how [K+], [H2PO4-], and [HPO4-2] are related. ([H+] and [OH-], although they are charged species, can be neglected for charge balance when compared to [K+], [H2PO4-], and [HPO4-2])

b. (10 pts) At pH 7.20 what is the ratio of [H2PO4-] to [HPO4-2]?

c. (20 pts) Combine your findings from parts a and b to determine how many mL of 1 M KOH must be added. {An important hint: By mass balance, the total molar concentration of phosphate species at pH 7.2 must equal the total molar concentration of phosphate species that was added from H3PO4.}

PLEASE ANSWER ALL PARTS, I NEED HELP. THANKS IN ADVANCE!

Calculate the pH at the equivalence point if 25.00 mL of 0.010 Mbarbituric acid (HC4H3N2O3) is titrated with 0.20 M NaOH. (Kabarbituric acid = 1.0 x 10^-5).


I calculated the pH to be 5.3. But I'm not sure which is why Iam asking.


25.0 mL x 0.010 M = 0.25 mmol Bacid

25.0 mL x 0.020 M = 0.50 mmol NaOH


Bacid NaOH

B 0.25 0

C 0.5-0.25 0.5

A 0.25 0.5


pKa = -log [Ka]

= -log (1.0 x 10^-5)

= 5


pH = pKa + log Base/Acid

pH = 5 + log 0.5/0.25

pH = 5 + 0.3010

pH = 5.3

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