Answer:
The percentage of molecules that are ionized is 1,1%
Explanation:
For the reaction:
C₆H₅COOH ⇌ C₆H₅COO⁻ + H⁺. The ka, 10^(-4,2), is defined as:
[tex]10^{-4,2} =\frac{ [C_6H_5COO^-][H^+]}{[C_6H_5COOH]}[/tex]
The molar concentrations in equilibrium are:
[C₆H₅COOH] = 0,57-x
[C₆H₅COO⁻] = x
[H⁺] = x
Replacing:
[tex]10^{-4,2} =\frac{ [x][x]}{[0,57-x]}[/tex]
3,6x10⁻⁵-6,3x10⁻⁵x = x²
0 = x² + 6,3x10⁻⁵x - 3,6x10⁻⁵
Resolving for x:
x = -0.006 Wrong answer, there are not negative concentrations
x = 0,006 Real answer
That means that:
[C₆H₅COOH] = 0,57-0,006 = 0,564M
[C₆H₅COO⁻] = 0,006M
The percentage of molecules ionized is [C₆H₅COO⁻]/initial[C₆H₅COOH], thus:
0,006M/0,57M×100 = 1,1%
I hope it helps
