A buffer solution is 0.20 M in HPO42- ions and 0.l0 M in H2PO4- ions. What is the H3O+ ion concentration of the solution? The ionization constants for H3PO4 are as follows: Ka1 = 7.5 ´ 10 -3, Ka2 = 6.2 ´ 10-8, Ka3 = 1 ´ 10-12.

a. 3.7 ´ 10-3 M
b. 3.7 ´ 10-4 M
c. 3.1 ´ 10-8 M
d. 3.1 ´ 10-9 M
e. 5 ´ 10-13 M

can anyone plz help with this questions

To determine the H3O+ ion concentration of the buffer solution, we need to consider the pKa values of H3PO4 and the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is given by:

pH = pKa + log([A-] / [HA])

Where:
pH = the acidity of the solution
pKa = the negative logarithm of the acid dissociation constant (Ka)
[A-] = concentration of the conjugate base
[HA] = concentration of the acid

In this case, H2PO4- is the conjugate base and HPO42- is the acid. The pKa values represent the degree of ionization of the acid.

First, let's find the pKa values for H3PO4.
pKa1 = -log(Ka1) = -log(7.5 x 10^-3) = 2.12
pKa2 = -log(Ka2) = -log(6.2 x 10^-8) = 7.21
pKa3 = -log(Ka3) = -log(1 x 10^-12) = 12

Now, let's calculate the pH using the Henderson-Hasselbalch equation.
pH = pKa + log([A-] / [HA])

For the given buffer solution:
[A-] = H2PO4- concentration = 0.10 M
[HA] = HPO42- concentration = 0.20 M

pH = pKa + log(0.10 / 0.20)
pH = pKa + log(0.5)
pH = pKa + (-0.30)

Since we want to find the H3O+ ion concentration, we can use the relationship:
[H3O+] = 10^(-pH)

Substituting the pH value into the equation:
[H3O+] = 10^(-(pKa - 0.30))

[H3O+] = 10^(-(2.12 - 0.30))
[H3O+] = 10^(-1.82)

[H3O+] ≈ 1.7 x 10^(-2)

Therefore, the H3O+ ion concentration of the solution is approximately 1.7 x 10^(-2) M.

None of the provided options match the calculated concentration.