Question

A student prepares a aqueous solution of trimethylacetic acid . Calculate the fraction of trimethylacetic 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

The question is incomplete, here is the complete question:

A student prepares a 0.21 mM aqueous solution of trimethylacetic acid. Calculate the fraction of trimethylacetic 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 3 significant digits.

Answer: The percent of dissociation of trimethylacetic acid is 19.5 %

Explanation:

We are given:

Concentration of trimethylacetic acid = 0.21 mM  = 0.00021 M    (Conversion factor:  1 mole = 1000 millimoles )

The chemical equation for the dissociation of trimethylacetic acid follows:

                     $C_4H_9COOH\rightleftharpoons C_4H_9COO^-+H^+$

Initial:                   0.00021

At eqllm:             0.00021-x              x            x

The expression of $K_a$ for above equation follows:

$K_a=\frac{[C_4H_9COO^-][H^+]}{[C_4H_9COOH]}$

We know that:

$K_a\text{ for }C_4H_9COOH=1\times 10^{-5}$

Putting values in above expression, we get:

$1\times 10^{-5}=\frac{x\times x}{(0.00021-x))}\\\\x=0.000041,-0.000051$

Neglecting the negative value of 'x' because concentration cannot be negative.

To calculate the fraction of dissociation, we use the equation:

$\text{Percent of dissociation}=\frac{[H^+]}{[C_4H_9COOH]}\times 100$

Putting values in above equation, we get:

$\text{Percent of dissociation}=\frac{0.000041}{0.00021}\times 100\\\\\text{Percent of dissociation of trimethylacetic acid}=19.5%$

Hence, the percent of dissociation of trimethylacetic acid is 19.5 %