Question

Review Intelligent Tutoring Question 16.082 Your answer is partially correct. Try again. A 0.15 M solution of an acid found in milk has a pH of 2.80. What is the percentage ionization of the acid? x Percent ionization = 1.06 What is the value of Ka for the acid? ka = 1.67E-5 Click if you would like to Show Work for this question: Open Show Work SHOW HINT LINK TO TEXT Question Attempts: 6 of 10 used SAVE FOR LATER SUBMIT ANSWER

Answer 1

Let a weak acid ionizes as

HA + H2O $\to$ H3O+  + A-

let concentration = C

Degree of ionization = x

ICE table is

Conentration	HA	H3O+	A
Initial	c	0	0
Change	-cx	+cx	+cx
Equilibrium	c - cx	cx	cx

Now,

[H3O+] = cx

pH = - log[ H3O+]

Given, pH = 2.80

Then, [H3O+] = 10-2.80 = 0.00158

Concentration of acid (c) = 0.15 M.

So, Cx = 0.00158

Or, 0.15x = 0.00158

Or, x = (0.00158/0.15) = 0.011 (2 significant figures)

Percent ionization of the acid

= 0.011×100 = 1.1 % .

Ka =

cx xсx с — ст

Or, Ka = $\frac{cx^{2}}{1- x}$

As, x <<1

Ka = cx2

Or, Ka = 0.15× (0.011)2 = 1.67 ×10-5