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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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ndixit1107 ndixit1107
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9 years ago
Another buffer found in blood is based on the equilibrium between dihydrogen phosphate and monohydrogen phosphate. The reaction is shown below:

H2PO4- + H2O ----> H3O+ + HPO42-

If the pH of a blood sample is 7.10, what would you calculate as the ratio of [H2SO4-] to [HPO42-]? (Ka1= 7.5*10^-3, Ka2= 6.2*10^-8, Ka3= 3.6*10^-13)
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#1
9 years ago
If I remember correctly, that equilibrium is for Ka2, so here's your Ka expression:

Ka2 = 6.2*10^-8 = ( [H+][HPO42-] ) / [H2PO4-]

So rearrange this to get the ratio of [H2SO4-] to [HPO42-]

[H2SO4-] / [HPO42-] = [H+] / 6.2*10^-8

Since you have pH, you can solve for [H+]:

pH = -log[H+]

So 10^-7.22 = [H+]

PS: I don't have a calculator with me, but just plug that number in and you'll have the ratio.
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#2
Educator
9 years ago
I don't think that's right Upwards Arrow Undecided
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