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.
1 Answer | Add Yours
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
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.
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes