Match the effect of addition of 0.1 M KOH to 0.1 M, 50 ml `H_3PO_4 Ka_1, Ka_2, Ka_3` are the I,II, III ionisation constant of `H_3PO_4` :
`{:("Column-I","Column-II"),((A)"75 ml of KOH", (p)pH=P^(Ka_1)),((B)"25 ml of KOH",(q)pH=P^(Ka_2)),((C )"150 ml of KOH",(r)pH=(P^(Ka_2)+P^(Ka_1))/2),((D)"100 ml of KOH",(s)pH=7+1/2[P^(Ka_3)+ logC]),(,(t)pOH=7-1/2[P^(Ka_3)+logC]):}`

Answer

A-q, B-p , C-s,t , D-r

Explanation

At 75 mL of `KOH^(-)`
2.5 mmol of `KH_2PO_4` and 2.5 mmol of `K_2HPO_4` are present in a solution.So using `pH=pKa_2+ log (["salt"])/(["acid"])`
But [salt]=[acid] hence pH=`pKa_2`
At 25 ml of `KOH-`
2.5 mmol of `KH_2PO_4` and 2.5 mmol of `H_3PO_4` are present in a solution.So using `pH=pKa_1+log(["salt"])/(["acid"])`
But [salt]=[acid] hence pH=`pKa_1`
At 150 ml of `KOH-`
`K_3PO_4` salt are present in a solution due to which salt hydrolysis will take place hence,
`pH=7+1/2[p^k a_3+ log C]`
At 100 ml of KOH-
`K_2HPO_4` are present in a solution which work as amphoteric in a solution so, `pH=(P^Ka_2+P^Ka_3)/2`