Calculate the pH of a weak acid which dissociates as AH ⇌ A- + H+ ( like CH3COOH ⇌ CH3COO- + H+ or HNO2 ⇌ NO2- + H+ ) knowing that the initial concentrations are [AH]0 0.97 M and the acid dissociation constant, Ka , is 7.23e-7

The pH is 3.08

Explanation:

Step 1:

The equation for the reaction.

AH ⇌ A- + H+

Step 2:

Data obtained from the question.

Initial concentration of AH, [AH] = 0.97 M

Dissociation constant, Ka = 7.23x10^-7

Step 3:

Determination of the concentration of A-, [A-], the concentration of H+, [H+] and the concentration of AH, [AH] after the reaction. This is illustrated below:

Before the reaction:

[AH] = 0.97 M

[A-] = 0

[H+] = 0

During the reaction:

[AH] = -y

[A-] = +y

[H+] = +y

After the reaction

[AH] = 0.97 - y

[A-] = y

[H+] = y

Step 4:

Obtaining the value of y. This is illustrated below:

Ka = [A-] [H+] / [AH]

Ka = 7.23x10^-7

[AH] = 0.97

[A-] = y

[H+] = y

Ka = [A-] [H+] / [AH]

7.23x10^-7 = y x y / 0.97

Cross multiply to express in linear form

7.23x10^-7 x 0.97 = y^2

7.0131x10^-7 = y^2

Take the square root of both side

y = √(7.0131x10^-7)

y = 8.37x10^-4

Therefore [H+] = y = 8.37x10^-4 M

Step 5:

Determination of the pH.

pH = - log [H+]

[H+] = 8.37x10^-4 M

pH = - log [H+]

pH = - log 8.37x10^-4

pH = 3.08

the answer is b & c i believe : )

the answers would be

the first one and the second one