Compare the two quantities based on the given condition: Ten (10) mL of solution X contains 0.100 M HCI and 0.100 M HA (pKa = 5) Ten (10) mL of solution Y contains 0.100 M HCl and 0.100 M HB (pKa = 7) Both solutions are titrated with 0.100 M NaOH. Volume of titrant required to reach the 2nd equivalence point in solution X Volume of titrant required to reach the 2nd equivalence point in solution Y. I > II I < II Cannot be compared I=II
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Compare the two quantities based on the given condition:
Ten (10) mL of solution X contains 0.100 M HCI and 0.100 M HA (pKa = 5)
Ten (10) mL of solution Y contains 0.100 M HCl and 0.100 M HB (pKa = 7)
Both solutions are titrated with 0.100 M NaOH.
- Volume of titrant required to reach the 2nd equivalence point in solution X
- Volume of titrant required to reach the 2nd equivalence point in solution Y.
I > II
I < II
Cannot be compared
I=II
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