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