Answers: 3
& 
----------------(i)
---------------(ii)![Ka_{1}= \frac{[H^{+} ][HA^{-} ]}{[H_{2}A ]}](/tpl/images/0570/7557/d943f.png)
![pH_{1} = -logKa_{1} + log\frac{[Base]}{[Acid]}](/tpl/images/0570/7557/2ee84.png)
![pH_{2}=-log Ka_{2} + log\frac{[Base]}{[Acid]} \\pH_{2}=-log Ka_{2} + log\frac{[A^{2-} ]}{[HA^{-} ]}\\\\7.22 = - logKa_{2} +log(1)\\\\Ka_{2}=10^{-7.22}= 6.02 X10^{-8}](/tpl/images/0570/7557/9dab2.png)
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For a certain diprotic acid, the pH at one half the volume to the first equivalence point is 4.15 an...
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