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3/4 of the way to second equivalence point of a diprotic acid/strong base titration3/4 of the way to second equivalence point of a diprotic acid/strong base titration
equivalence point of a weak base/strong acid titrationequivalence point of a weak base/strong acid titration
equivalence point of a strong acid/strong base titrationequivalence point of a strong acid/strong base titration
equivalence point of a weak acid/strong base titrationequivalence point of a weak acid/strong base titration
half-way to equivalence point of a weak acid/strong base titrationhalf-way to equivalence point of a weak acid/strong base titration

Please see below for answers

Explanation: Matching appropriate labels , we have

1)3/4 of the way to second equivalence point of a diprotic acid/strong base titration-- pH=pka₂

equivalence point of a weak base/strong acid titration=pH<7

equivalence point of a strong acid/strong base titration= pH=7

equivalence point of a weak acid/strong base titration=pH>7

half-way to equivalence point of a weak acid/strong base titration pH =pka

where

pH gives the measure of the amount of concentration of hydrogen ions in an aqueous solution.

pKa  is known as acid dissociation constant which explains the equilibrum at which a chemical species can give out or receive proton

pka₂ is the acid dissociation constant for the second ionization energy.

d.) circle motion requires a force to give it a centripetal acceleration.

i'm assuming that it's c.) red bank..