Phosphate can act as buffer: H2PO4- HPO4^2- + H+
The pKa of this equilibrium is 7.21. What is the ratio of H2PO4-
to HPO4^2- at pH 7.0?
A. 1.62
B. 0.62
C. 50.47
D. 0.019
E. 0.21
The HH equation calculates this directly.
To solve this problem, we need to use the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log([A-]/[HA])
Where pH is the desired pH value, pKa is the acidity constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, the conjugate base is HPO4^2- and the acid is H2PO4-. We want to find the ratio of [H2PO4-] to [HPO4^2-] at pH 7.0, which means we can substitute these values into the Henderson-Hasselbalch equation:
7.0 = 7.21 + log([HPO4^2-]/[H2PO4-])
Now we can rearrange this equation to solve for the ratio [HPO4^2-]/[H2PO4-]:
log([HPO4^2-]/[H2PO4-]) = 7.0 - 7.21
log([HPO4^2-]/[H2PO4-]) = -0.21
Next, we can use the property of logarithms that states:
log(a/b) = log(a) - log(b)
Applying this to our equation, we get:
log([HPO4^2-]) - log([H2PO4-]) = -0.21
Now we can convert the logarithm equation into exponent form:
[HPO4^2-]/[H2PO4-] = 10^(-0.21)
Using a calculator, we can find that 10^(-0.21) is approximately 0.621.
Therefore, the ratio of [H2PO4-] to [HPO4^2-] at pH 7.0 is 0.621.
So, the correct answer is B. 0.62.