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Urine is buffered by a NaH2PO4 - Na2HPO4 buffer. If the pH of a urine specimen is 6.6,...

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bobby9901 | Student, College Freshman | (Level 1) Salutatorian

Posted April 18, 2012 at 6:44 PM via web

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Urine is buffered by a NaH2PO4 - Na2HPO4 buffer. If the pH of a urine specimen is 6.6, what is the ratio of [HPO4^2‒] / [H2PO4^‒]?

For phosphoric acid, Ka2 = 6.2 x 10^‒8.

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mlsiasebs | College Teacher | (Level 1) Associate Educator

Posted April 19, 2012 at 2:25 PM (Answer #1)

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Since we have a buffer solution, we can use the Henderson-Hasselbalch equation to determine the ratio.

pH = pKa + log (base/acid)

We know the pH of the solution, we know the Ka so we can find the pKa and we can solve for the base/acid ratio.  To determine which one is the acid and which one is the base, we need to look at the difference between the ions. 

HPO4^2- is accepting a proton to become H2PO4- so it is acting as a base.  Since H2PO4- is donating a proton, it is the acid.  In the Henderson-Hasselbalch equation, the concentration of the base (i.e. the A-) is on top and the concentration of the acid is on the bottom. 

To find the pKa we need to take the - log Ka which for this problem will be -log (6.2 x 10^-8) = 7.21

Therefore

6.6 = 7.21 + log (base/acid)

Since what we want to know is the ratio, we can solve for the (base/acid)

-0.6 = log (base/acid)

The opposite of log is 10^ so we can take both sides as the powers of 10

10^-0.6 = 10^(log(base/acid)

10^-0.6 = base/acid

0.25 = base/acid which is our ratio for the base to acid concentrations.

 

 

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