A buffer that consists of a mixture of NaH2PO4 and K2HPO4 has a pH of 6.73. The (HPO4)/(H2PO4) ratio in this buffer is what? If (k1,k2,and k3 for H3PO4 are 7.1x10^-3, 6.2x10^-8, and 4.5x10^-13 respectively)
Substitute the pH and pKa into
pH = pKa + log(base)/(acid) and solve for the ratio of (base)/(acid).
To determine the (HPO4)/(H2PO4) ratio in the buffer, we can utilize the Henderson-Hasselbalch equation, which relates the pH of a buffer to the ratio of its conjugate acid and base forms.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
In this equation, pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In the case of the phosphate buffer, the relevant acid and conjugate base forms are H2PO4- (acid) and HPO42- (conjugate base).
The dissociation of H3PO4 can be represented by the following equations:
H3PO4 ⇌ H+ + H2PO4- (k1)
H2PO4- ⇌ H+ + HPO42- (k2)
HPO42- ⇌ H+ + PO43- (k3)
Since we are given k1, k2, and k3, we can use these values to calculate pKa values for each equilibrium reaction.
pKa1 = -log(k1)
pKa2 = -log(k2)
pKa3 = -log(k3)
Now, we need to determine the concentrations of H2PO4- and HPO42- in the buffer solution. Let's assume:
[H2PO4-] = x
[HPO42-] = y
Since the buffer is neutral (pH = 6.73), we can set up the following equation using the Henderson-Hasselbalch equation:
6.73 = pKa2 + log(y/x)
Now, we can solve for the ratio (HPO4)/(H2PO4) by rearranging and solving the equation:
log(y/x) = 6.73 - pKa2
Taking the antilog of both sides of the equation, we get:
y/x = 10^(6.73 - pKa2)
Using the given value of pKa2 (6.2x10^-8), we can substitute it into the equation:
y/x = 10^(6.73 - 6.2x10^-8)
Calculating the right side of the equation gives us the ratio (HPO4)/(H2PO4) in the buffer.
Please note that the actual calculation depends on the specific values of [H2PO4-] and [HPO42-] which might be provided in the problem or need to be obtained from additional information.