Answer:
pH =2.685
Explanation:
mass of acetylsalicylic acid, m = 2 × 325 m g × (1 g / 1000 m g )
= 0.65 g
Volume V = 237mL
dissociation constant, Ka =3.3×10⁻⁴
molecular weight of acetylsalicylic acid = 180.1 g/mol
mass of acetylsalicylic acid, (HC₉H₇O₄)
= 0.65 / 180.1
= 0.0036mol
concentration of HC₉H₇O₄ in a 237 mL solution
M = 0.0036 / 237ML
= 0.015M
in a 237 mL solution HC₉H₇O₄ in water is
C₉H₇O₄⁻ ⇄ H⁺
Next, we show the changes in different phases that occur during the dissociation process. We use the value x as the concentration loss/gained during the dissociation process,
HC₉H₇O₄ C₉H₇O₄⁻ H⁺
Initial 0.015 0 0
change -x x x
equilibrium 0.015 -x x x
The equation for the dissociation constant Ka ,
[tex]K_a = \frac{[C_9H_7O_4^-[H^+]]}{[HC_9H_7O_4]} \\3.3 * 10^-^4 = \frac{x * x}{0.015 - x} \\4.95 * 10^-^6-3.3 * 10 ^-^4x = x^2\\[/tex]
using quadratic equation
x² + 3.3 * 10⁻⁴x - 4.95 * 10 ⁻⁶ = 0
x = 0.002066M
pH = -log[H⁺]
pH = -log[0.002066]
= 2.685
