Answer:
a) = 0.704%
b) = 1.30%
c) = 2.60%
Explanation:
Given that:
[tex]K_a[/tex]= [tex]1.34*10^{-5[/tex]
For Part A; where Concentration of A = 0.270 M
Percentage Ionization(∝) [tex]\alpha = \sqrt{\frac{K_a}{C} }[/tex]
[tex]\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }[/tex]
[tex]\alpha = \sqrt{4.9629*10^{-5}}[/tex]
[tex]\alpha = 7.044*10^{-3[/tex]
percentage% (∝) = [tex]7.044*10^{-3}*100[/tex]
= 0.704%
For Part B; where Concentration of B = [tex]7.84*10^{-2[/tex] M
[tex]\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }[/tex]
[tex]\alpha = \sqrt{1.709*10^{-4} }[/tex]
[tex]\alpha = 0.0130\\[/tex]
percentage% (∝) = 0.0130 × 100%
= 1.30%
For Part C; where Concentration of C= [tex]1.92*10^{-2} M[/tex]
[tex]\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }[/tex]
[tex]\alpha = \sqrt{6.802*10^{-4}}[/tex]
[tex]\alpha =0.02608[/tex]
percentage% (∝) = 0.02608 × 100%
= 2.60%
The percent ionization for propionic acid in each case are:
Percent ionization is the quantity of a weak acid that ionizes in a solution expressed as a percentage.
We can calculate the percent ionization using the following expression.
α% = √(Ka/Ca) × 100%
where,
α% = √(1.34 × 10⁻⁵/0.270) × 100% = 0.704%
α% = √(1.34 × 10⁻⁵/7.84 × 10⁻²) × 100% = 1.31 %
α% = √(1.34 × 10⁻⁵/1.97 × 10⁻²) × 100% = 2.61 %
The percent ionization for propionic acid in each case are:
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