Question

Another buffer found in blood is based on the equilibrium between dihydrogen phosphate and monohydrogen phosphate. The reaction is shown below:

H2PO4^-(aq) + H2O(l)----> H3O^+(aq) + HPO4^-2(aq)

If the pH of a blood sample was 7.10, what would you calculate as the ratio of (H2PO4^-) to (HPO4^-2)?

(Ka1 = 7.5*10^-3, Ka2 = 6.2*10^-8, Ka3 = 3.6*10^-13)

Answer 1

Ans- firstly, we have to find the conc. of H⁺ at pH of 7.4. ,hence 10^7.4,= 4E-8.
[H⁺] = 4E-8
so, conc. of H₃PO₄: So we know that Ka of this is 7.5E-3, hence it's Ka expression would be: 7.5E-3 = [H⁺][HPO4]/[H2PO4].

, remembering that [H+] = 4E-8, and that if we were to write out the ICE table, we'd write x₂/[H2PO4]=7.5E-3
7.5E-3 = [4E-8]^2/[H3PO4]
therefore:
[4E-8]^2/7.5E-3 = [H3PO4]
[H3PO4]= 2.13E-13

we just found for H3PO4 is actually H2PO4. But now, because we're wanting to find HPO4 we use the Ka for HPO4 = 3.6E-13.
therefore: 3.6E-13 = [H⁺][HPO4]/[H2PO4]
3.6E-13 = (4E-8)[HPO4]/(2.13E-13)
((3.6E-13)x(2.13E-13)/(4E-8))=[HPO4]
[HPO4] = 2.52E-18

Now we find the REAL H2PO4 concentration, by the same steps as above; subbing it's Ka into the equation:
6.2E-8 = [H+][HPO4]/[H2PO4]
[H2PO4] = [H+][HPO4]/6.2E-8
[H2PO4] = (4E-8)x(2.52E-18)/6.2E-8
[H2PO4] = 1.6E-18

Now ratio between H2PO4 and HPO4, you divide the concentration of H2PO4 by HPO4:
[H2PO4]/[HPO4]
=1.6E-18 / 2.52E-18
=0.63

please recheck the calculation but method is correct