From the list below, pick an acid, along with its conjugate base, that can be used to make a pH 7.40 buffer.
H3PO4 Ka = 6.9 x 10-3 H2PO4- Ka = 6.4 x 10-8 HPO42- Ka = 4.8 x 10-13
H3PO4 Ka values :
Ka1 = 6.9 x 10^-3 ; pKa1 = 2.16
H3PO4 <---------> H+ + H2PO4-
H2PO4- <----------> H+ + HPO42-
HPO42- <-------------> H+ + PO43-
Ka2 = 6.4 x 10^-8 ; pKa2 = 7.19
Ka3 = 4.8 x 10^-13 ; pKa3 = 12.32
here the given pH of buffer = 7.40
for best buffer : pH = pKa +/- 1
here pKa2 value is close the given pH .
so for prepare pH of 7.40 buffer use
buffer pair : H2PO4- and HPO42-
From the list below, pick an acid, along with its conjugate base, that can be used...
a) From the list below, pick an acid, along with its conjugate base, that can be used to make a pH 7.40 buffer. H3PO4 Ka = 6.9 x 10^-3 H2PO4- Ka = 6.4 x 10^-8 HPO42- Ka = 4.8 x 10^-13 b) If you have 0.100 M solutions of the acid and the conjugate base in the system that you picked and there was 20.0 mL of the base, how many mL of the acid would you need to make...
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