Respuesta :
Answer:
pH = 2.76
Explanation:
Given that:
[tex]K_{a}=3.3\times 10^{-4}[/tex]
Given that:
Mass = 440 mg = 0.44 g (1 mg = 0.001 g)
Molar mass = 180.158 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{0.44\ g}{180.158\ g/mol}[/tex]
[tex]Moles= 0.00244\ moles[/tex]
Given Volume = 270 mL = 0.27 L ( 1 mL = 0.001 L)
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Molarity=\frac{0.00244}{0.27}[/tex]
Concentration = 0.00904 M
Consider the ICE take for the dissociation of acetic acid as:
HC₉H₇O₄ ⇄ H⁺ + C₉H₇O₄⁻
At t=0 0.00904 - -
At t =equilibrium (0.00904-x) x x
The expression for dissociation constant of acetic acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {C_9H_7O_4}^- \right ]}{[HC_9H_7O_4]}[/tex]
[tex]3.3\times 10^{-4}=\frac {x^2}{0.00904-x}[/tex]
x is very small, so (0.00904 - x) ≅ 0.00904
Solving for x, we get:
x = 1.7272×10⁻³ M
pH = -log[H⁺] = -log(1.7272×10⁻³) = 2.76
