Question

A student prepares a aqueous solution of 4chlorobutanoic acid . Calculate the fraction of -chlorobutanoic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource. Round your answer to significant digits.

Answer 1

Incomplete question: The concentration of the aqueous solution is 0.65 mM

The percentage of dissociation is 19.23%

Step-by-step explanation:

The pKa of the 4-chlorobutanoic acid is 4.53, thus:

`pKa=-logKa\\Ka=10^(-4.53) =2.95x10^(-5)`

The initial concentration is 0.65 mM = 6.5x10⁻⁴M

The reaction is:

HA + H₂O = H₃O⁺ + A⁻

I 6.5x10⁻⁴ 0 0

C -x +x +x

E 6.5x10⁻⁴ -x x x

The Ka is:

`Ka=([H_(3)O^(+) ][A^(-)] )/([HA]) \\2.95x10^(-5) =(x*x)/(6.5x10^(-4)-x ) \\x^(2) +2.95x10^(-5)x-1.92x10^(-8) =0\\x=1.25x10^(-4) M=[A^(-) ]`

The percentage of dissociated is:

`P=([A^(-)] )/([HA]) *100=(1.25x10^(-4) )/(6.5x10^(-4) ) *100=19.23`%