Answer:
There is more A⁻ than HA in the solution
Step-by-step explanation:
The equation for the ionization of a weak acid is
HA + H₂O ⇌H₃O⁺ + OH⁻
When HA and A⁻ are present in comparable amounts, we can use the Henderson-Hasselbalch equation:
![\begin{array}{rcl}\text{pH} &=& \text{pK}_{\text{a}} + \log\frac{[\text{A}^(-)]}{\text{[HA]}}\\\\5.5 & = & 4.76 + \log\frac{[\text{A}^(-)]}{\text{[HA]}}\\\\0.74 & = & \log\frac{[\text{A}^(-)]}{\text{[HA]}}\\\end{array}](https://img.qammunity.org/2020/formulas/chemistry/college/j7136jmz9pmyfqugrslaupmdqdg5htgigq.png)
0.74 is a positive number, and a number must be greater than one for its logarithm to be positive. That is,
![\begin{array}{rcl}\frac{[\text{A}^(-)]}{\text{[HA]}} & > & 1\\\\\textbf{[A]}^(-) & > & \textbf{[HA]}\\\end{array}](https://img.qammunity.org/2020/formulas/chemistry/college/maumsh6p3gapgh1p5t0qka45u09gn5xmj4.png)