Determine the percent ionization of a solution of boric acid that is 51 mM (Ka = 5.4 × 10 w/exponent -10).?
Answer:
Boric acid reacts with water as follows
B(OH)₃ + H₂O ←→ B(OH)₄⁻ + H⁺
with
Ka = [B(OH)₄⁻] · [H⁺] / [B(OH)₃]
[X] is the concentration in M
The ion concentrations in a solution of boric acid of apparent concentration c, which is ionized to the degree α are:
[B(OH)₄⁻] = [H⁺] = α · c
[B(OH)₃] = (1 - α) · c
Hence:
Ka = α² · c² / (1 - α) · c
<=>
Ka · (1 - α) = α² · c
<=>
α² · (c/Ka) + α - 1 = 0
The quadratic equation has two solutions:
α = (Ka/ c)·(-1 ± √(1 + 4 (c/Ka)) / 2
Only the positive solution makes sense thus:
α = (5.4×10^-10 / 0.051)·(-1 + √(1 + 4 (0.051 / 5.4×10^-10)) / 2
= 1.03×10^-4 = 0.0103%
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