a)
moles of K2HPO4- = 52.25 / 174.17 = 0.3
concentration of HPO42- = 0.3 x 400 / 1000 = 0.12 M
HPO42- + H2OÂ Â ----------------> H2PO4- + OH-
0.12Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â
0
0.12 -
x                                             Â
x           Â
x
Kb2 = x^2 / 0.12 - x
1 x 10^-14 / 6.2×10–8 = x^2 / 0.12 - x
x = 1.39 x 10^-4
[OH-] = 1.39 x 10^-4 M
pOH = -log [OH-] = -log(1.39 x 10^-4)
        = 3.86
pH = 10.14
pH = pKa2 + log [salt / acid]
7.3 = 7.21 + log [HPO42-]/[H2PO4-]
[HPO42-]/[H2PO4-] = 1.23
[H2PO4-]/[HPO42-] = 0.813