Question

*Remember to submit each part separately to avoid losing your work*

A 0.100 M solution of an unknown acid, HA, is 3.87% ionized.

Complete the following table, but instead of using the variable x, calculate the number values
of each quantity and report these values to the correct number of significant figures.

Fill in all of the cells.
Include signs in the Change column to indicate a gain or loss of concentration.
Do not include any units in your answers, M is assumed.

To enter values in scientific notation follow the convention: 0.00753 = 7.53E-3

	[HA]	[H+]	[A-]
Initial
Change
Equilibrium

K_a= What is the K_a of HA? Calculate your answer to three sig figs.

Now calculate the pH of the solution of HA

pH=

Answer 1

The egiven acid HA is ionised by 3.87%. The means it is a weak acid . Now the dissociation of weak acicd is as-

HA + H2O ------> H3O+ + A-

Where HA = Weak acid A- = conjugate base.

The pH of weak acid is calculated by the formula

pH = pKa + log [Base] / [Acid]

This concentration is the concentration at equilibrium

Now to find this concentration by formring ICE table

The initial Acid taken = 0.100 M

Ionised = 3.87% = 0.100 M * 3.87% = 0.00387‬ M

That means amount of A- formed after ionisation = 0.00387‬ M

And the amount of HA left = 0.100 M - 0.00387‬ M = 0.09613‬ M

Again from the equation, both H3O+ and A- formed are equimolar

Thus

	[HA]	[H+]	[A-]
Initial	0.100	0	0
Change	- 0.00387‬ M	+ 0.00387‬ M	+ 0.00387‬ M
Equilibrium	0.09613‬ M	0.00387‬ M	0.00387‬ M

Now Ka = [H+] * [A-] / [HA]

= 0.00387‬ M * 0.00387‬ M / 0.09613‬ M

= 1.55 * 10^-4

Thus pKa = - log Ka = - log (1.55 * 10^-4)

= 4 - log 1.55

= 3.809

Hence pH = pKa + log [Base] / [Acid]

= 3.809 + log [0.00387‬ M] / [0.09613‬ M]

= 2.41