Answer:
pH = 8.84
Explanation:
The reaction of HA + NaOH is:
HA + NaOH → H₂O + Na⁺ + A⁻
20.00mL of 0.150M HA are:
0.020L × (0.150mol / L) = 3.00x10⁻³ moles of HA
Moles of NaOH are:
0.015L × (0.150mol / L) = 3.00x10⁻³ moles of NaOH
That means all HA reacts producing 3.00x10⁻³ moles A⁻. In 35mL of total volume: 3.00x10⁻³ moles A⁻ / 0.035L = 0.0857M.
The equilibrium of A⁻ in water is:
A⁻ + H₂O ⇄ HA + OH⁻
Where Kb is (Kw / Ka = 1.00x10⁻¹⁴ / 1.80x10⁻⁵ = 5.56x10⁻¹⁰) defined as:
Kb = [HA] [OH⁻] / [A⁻]
Equilibrium concentrations are:
[A⁻] = 0.0857M - X
[HA] = X
[OH⁻] = X
Replacing in Kb formula:
5.56x10⁻¹⁰ = [X] [X] / [0.0857M - X]
4.77x10⁻¹¹ - 5.56x10⁻¹⁰X = X²
0 = X² + 5.56x10⁻¹⁰X - 4.77x10⁻¹¹
Solving for x:
X = -6.907x10⁻⁶ M → False answer → There is no negative concentrations
X = 6.906x10⁻⁶ M → Right answer
As X = [OH⁻] and pOH = -log [OH⁻], pOH =-log 6.906x10⁻⁶ M = 5.16
As 14 = pH + pOH
pH = 14 - 5.16
<em>pH = 8.84</em>
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