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When 1.00 L of 0.45 M acetic acid (pKa = 4.74) is mixed with the exact...

When 1.00 L of 0.45 M acetic acid (pKa = 4.74) is mixed with the exact volume of 0.55 M NaOH required to convert the acid to its conjugate base, at the endpoint the solution will have a:

A. pH < pKa

B. pH > 7.00

C. pH = pKa

D. pH = 7.00

E. pH < 7.00

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Answer #1

moles of NaOH = molarity * volume

= 0.55 * 1 (same volume)

= 0.55

moles of acetic acid = 0.45 * 1

= 0.45

remaning moles of NaOH = 0.55 - 0.45

= 0.10

so, pH will be greater than 7.

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