What are the major species present in 0.250 M solutions of...

What are the major species present in 0.250 $M$ solutions of each of the following acids? Calculate the pH of each of these solutions.
a. $\mathrm{HNO}_{2}$
b. $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)$

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Key Concepts

Acid Dissociation Equilibrium

This concept explains how weak acids only partially donate protons to water, establishing an equilibrium between the undissociated acid and its dissociation products (H? and A?). It is fundamental for understanding the composition of species in solution for weak acids, as the equilibrium state determines the concentrations of all species present.

Equilibrium Constant (Ka)

The acid dissociation constant, Ka, is a measure of the extent to which a weak acid ionizes in solution. It is defined by the ratio of the concentrations of the products (H? and A?) to the concentration of the undissociated acid (HA). This constant is crucial for calculating the concentrations of all species at equilibrium and influences the pH of the solution.

pH Calculation

pH is the logarithmic measure of the hydrogen ion concentration in a solution, expressed as pH = -log[H?]. Calculating the pH of a weak acid solution typically involves setting up the equilibrium expression associated with the acid dissociation and solving for [H?], either exactly or with appropriate approximations when the degree of dissociation is small.

Transcript

00:05 For a, we have hno2, which is a weak acid.

00:11 And we're asked, what are the major species present and the ph of a .250 molar solution? the major species present will be, since it is a weak acid, hno2 and h2o.

00:30 And since it is a weak acid, we need to set up a nice table.

00:36 H02.

00:38 And h2o, in equilibrium with h3l plus, and n02 minus.

00:48 We're told the initial molarity is 0 .250 molar.

00:51 Let's complete our ice table.

01:01 K -a expression.

01:11 We're given the k -a value as 4 .0.

01:17 We can look this up.

01:18 Let's send of the negative 4.

01:19 Is equal to x, x, and 0 .250 minus x.

01:28 Since this is a weak acid and avoid solving the quadratic, we can assume that the denominator, 0 .250 minus x, x is infinitely small, and this would just be equal to 0 .250.

01:41 So our equation will simplify to 4 .0 times 10 of the negative 4, x squared in the numerator, 0 .250 and the denominator.

01:50 Solving for x will yield 0 .010.

02:02 Therefore, going back to our ice table, the h -30 plus molarity is 0 .010 molar.

02:10 And based on this, our ph would be negative log.

02:14 010.

02:15 And we'll find that the ph will be equal to 2 .00.

02:27 We have c .h3 c .o .2 .h.

02:34 Which is a weak acid...