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Title: Question Regarding Buffers and pH Post by: ndixit1107 on Apr 8, 2015 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) Title: Re: Question Regarding Buffers and pH Post by: psyche360 on Apr 9, 2015 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. Title: Re: Question Regarding Buffers and pH Post by: bio_man on Apr 9, 2015 I don't think that's right ^^ :-\
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