Question

Aspirin is a weak organic acid whose molecular formula is HC9H7O4. An aqueous solution of aspirin is prepared by dissolving 3.60 g/L. The pH of this solution is found to be 2.6. Calculate Ka for aspirin. (atomic mass: C

Answer 1

The Ka for aspirin is approximately 3.0 x 10^-4.

Step-by-step explanation:

To calculate Ka for aspirin, we need to understand that aspirin (acetylsalicylic acid, CH3CO₂C6H4CO₂H) acts as a weak acid in water. The primary acidic functional group in aspirin is the carboxylic acid group (-CO₂H). The pH of the aspirin solution is given as 2.6.

Using the pH value, we can calculate the concentration of hydronium ions in the solution. From the concentration of hydronium ions, we can then determine the concentration of aspirin that underwent ionization. Lastly, we can calculate Ka using the concentration of reactants and products in the ionization equation.

Using this approach, we can find that the Ka for aspirin is approximately 3.0 x 10-4.

Answer 2

Ka = 3.50x10⁻⁴

Step-by-step explanation:

First, we need to convert the unit of 3.60 g/L to mol/L:

`C_{C_(9)H_(8)O_(4)} = 3.60 (g)/(L)*(1 mol)/(180.16 g) = 0.0200 mol/L`

The reaction dissociation of aspirin in water is:

C₉H₈O₄ + H₂O ⇄ C₉H₇O₄⁻ + H₃O⁺

0.02 - x x x

The constant of the above reaction is:

`Ka = ([C_(9)H_(7)O_(4)^(-)][H_(3)O^(+)])/([C_(9)H_(8)O_(4)])`

`Ka = (x^(2))/(0.02 - x)`

To find Ka we need to find the value of x. We know that pH = 2.6 so:

`pH = -log[H_(3)O^(+)]`

`2.6 = -log(x)`

`x = 2.51 \cdot 10^(-3) M = [H_(3)O^(+)] = [C_(9)H_(7)O_(4)^(-)]`

Now, the concentration of C₉H₈O₄ is:

`C_{C_(9)H_(8)O_(4)} = 0.02 - 2.51 \cdot 10^(-3) = 0.018 M`

Finally, Ka is:

`Ka = ([C_(9)H_(7)O_(4)^(-)][H_(3)O^(+)])/([C_(9)H_(8)O_(4)]) = ((2.51 \cdot 10^(-3))^(2))/(0.018) = 3.50 \cdot 10^(-4)`

Therefore, the Ka of aspirin is 3.50x10⁻⁴.

I hope it helps you!