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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 ofKOH",(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]):} |
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Answer» 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` |
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