Answer:
a. pH is 2,88
b. pH is 8,87
Explanation:
a. Acetic acid (HC₂H₃O₂) dissociates in water thus:
HC₂H₃O₂ ⇄ C₂H₃O₂⁻ + H⁺; Ka = 1,8x10⁻⁵
ka = [C₂H₃O₂⁻] [H⁺] / [HC₂H₃O₂] (1)
When you have a solution of 0,10M HC₂H₃O₂, the concentrations in equilibrium are:
[HC₂H₃O₂] = 0,10M - x
[C₂H₃O₂⁻] = x
[H⁺] = x
Replacing in (1)
1,8x10⁻⁵ = x² / 0,10 - x
1,8x10⁻⁶ - 1,8x10⁻⁵x = x²
x²+ 1,8x10⁻⁵x - 1,8x10⁻⁶ = 0
Solving for x:
x = - 0,00135067 -No physical sense, there are not negative concentrations-
x = 0,00133267 -Real answer-
As x = [H⁺] = 0,00133267
pH = -log [H⁺]
Thus, pH is 2,88
b. Sodium acetate (NaC₂H₃O₂) dissociates in water thus:
NaC₂H₃O₂ + H₂O ⇄ HC₂H₃O₂ + OH⁻ + Na⁺; Kb = 5,6x10⁻¹⁰
ka = [HC₂H₃O₂] [OH⁻] / [NaC₂H₃O₂] (1)
When you have a solution of 0,10M NaC₂H₃O₂, the concentrations in equilibrium are:
[NaC₂H₃O₂] = 0,10M - x
[HC₂H₃O₂] = x
[OH⁻] = x
Replacing in (1)
5,6x10⁻¹⁰ = x² / 0,10 - x
5,6x10⁻¹¹ - 5,6x10⁻¹⁰x = x²
x²+ 5,6x10⁻¹⁰x - 5,6x10⁻¹¹ = 0
Solving for x:
x = -7,48359×10⁻⁶ -No physical sense, there are not negative concentrations-
x = 7,48303×10⁻⁶ -Real answer-
As x = [OH⁻] = 7,48303×10⁻⁶
pOH = -log [OH⁻]
pOH is 5,13. As 14 = pH + pOH
pH = 8,87
I hope it helps!
