Answer:
True
Explanation:
Given:
pKa of weak acid = 4.82
pH of buffer = 4.25
Now, put the values in Henderson-Hasselbalch equation
[tex]pH=p_{Ka} + log\frac{[Salt]}{[Acid]}[/tex]
[tex]4.25=4.82 + log\frac{[A^{-}]}{[HA]}[/tex]
[tex]log\frac{[A^{-}]}{[HA]} = -0.57[/tex]
[tex]\frac{[A^{-}]}{[HA]} = 0.269[/tex]
0.269 is less than 1.
therefore, [HA] is greater than [A-].
Hence, the given statement is true.
