Question

Another buffer found in blood is based on the equilibrium between dihydrogen phosphate and monohydrogen phosphate....

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)

Homework Answers

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

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