pH and Ka values of Weak Acids

AP Chem April 24, 2012

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04/19/23 2

Weak Acids: Calculation of Ka from pH

• Will need to use ICE skills for solving equilibrium problems.

• Because the concentration of the acid (reactant side) does NOT equal the concentration of the H+ ion (product side)

• There is far less than 100% ionization taking place.

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A student prepared a 0.10 M solution of formic acid (HCHO2).

A pH meter shows the pH = 2.38.

a. Calculate Ka for formic acid.

b. What percentage of the acid ionized in this 0.10 M solution?

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HCHO2 (aq) H+ (aq) + CHO2- (aq)

• First, let’s find the [H+] from the pH

• [H+] = 10(-2.38) • = 4.2 x 10-3 M• Great, Now for some ICE

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HCHO2 H+ CHO2-

I 0.10 M 0 0

C

E

HCHO2 (aq) H+ (aq) + CHO2- (aq)

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HCHO2 H+ CHO2-

I 0.10 M 0 0

C -4.2 x 10-3 M +4.2 x 10-3 M +4.2 x 10-3 M

E 0.10 -4.2 x 10-3 M = 0.0958

4.2 x 10-3 M 4.2 x 10-3 M

Assumed from the pH [H+]

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So, now for the Ka calculation:

= 1.8 x 10-4

Is our answer reasonable? Yes, Ka values for weak acids are usually between 10-3

and 10-10.

2

2

HCHO

CHOHKa

0958.0

)102.4)(102.4( 33 aK

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And what about the percent ionization stuff?

• Formula to use:

% = 4.2 x 10-3 x 100 = 4.2 % 0.10

100IonizationPercent

initial

equil

Acid

H

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Niacin, one of the B vitamins, has the following molecular structure:

                    A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution? (b) What is the acid-dissociation constant, Ka, for niacin?

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Niacin Problem #1

• pH = 3.26 [H+] = ?• [H+] = 10-3.26 = 5.50 x 10-4 M• Percent Ionization = [H+]equilibrium x 100

[Acid]Initial

= 5.50 x 10-4 M / 0.02 M x 100 = 2.7 %

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Solution to Niacin Problem

• Ka = [H+] [ niacin ion-] [niacin] Niacin H+ niacin ionI 0.02 0 0C - 5.50 x 10-4 + 5.50 x 10-4 + 5.50 x 10-4

E 0.02 - 5.50 x 10-4 5.50 x 10-4 5.50 x 10-4 • Ka = (5.50 x 10-4)2 = 1.55 x 10-5

0.019

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Using Ka to Calculate pH

• Similar to the approach we used in Chapter 15, sometimes using the quadratic equation to solve for the equilibrium concentrations. Once you know the equilibrium concentration of [H+], you can calculate the pH.

• Need to have Ka value and the initial concentration of the weak acid

• Start by writing equation and equilibrium-constant expression for the reaction.

• Let’s calculate the pH of a 0.30 M solution of acetic acid (HC2H3O2) at 250C.

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First step: Write the ionization equilibrium for acetic acid:

• HC2H3O2(aq) H+ (aq) + C2H3O2- (aq)

Second Step: Write the equilibrium-constant expression

• Ka = [H+ ] [C2H3O2- ] = 1.8 x 10-5

[HC2H3O2]

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Step 3: Set up an ICE calculation

HC2H3O2(aq) H+ (aq) + C2H3O2- (aq)

I 0.30 M 0 0

C -x M +x M +x M

E (0.30 – x) M x M x M

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Fourth Step: Substitute the equilibrium conc into expression.

• Ka = [H+ ] [C2H3O2- ] = 1.8 x 10-5

[HC2H3O2]• = (x) (x) = 1.8 x 10-5

(0.30 –x)Solve using quadratic equation: x = 2.3 x 10-3 M• Percent Ionization = [H+]equilibrium x 100 [Acid]Initial

% ionization = 0.0023 M x 100 = 0.77% 0.30 M

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• Calculate the pH of a 0.20 M solution of HCN (Refer to table 16.2 or Appendix D for the Ka value.)

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Solution

• HCN (aq) H+ (aq) + CN-(aq)• Ka = [H+ ] [CN-] = 4.9 x 10-10

[HCN]

I 0.20 M 0 0C -x M +x M +x ME 0.20 – x x M x M

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• (x) (x) = 4.9 x 10-10

(0.20 –x)Use quadratic equation to solve for x:

x2 = 4.9 x 10-10(0.20 – x)x2 + 4.9 x 10-10x – 9.8 x 10-11 = 0 x = 9.9 x 10-6 = [H+]pH = -log(9.9 x 10-6) pH = 5.00

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Second Niacin problem

The Ka for niacin is 1.6 x 10-5. What is the pH of a 0.010 M solution of niacin?

1st find the [H+] at equilibrium

Niacin H+ niacin ion

Initial 0.010 0 0

Change -x +x +x

Equilibrium 0.010-x x x

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• Ka = [H+] [niacin ion] = 1.6 x 10-5

[niacin]1.6 x 10-5 = x2 / (0.010-x)x2 + 1.6 x 10-5 x - 1.6 x 10-7 = 0x = 3.92 x 10 -4 = [H+]pH = -log(3.92 x 10 –4) pH = 3.41

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5.00 mL of 0.250 M HClO3 diluted to 50.0 mL; pH =?

6.1)025.0log(

H M025.0soln L 0.050

mol 00125.0

H 00125.0soln L 1

mol 250.0

mL 1000

L 1 mL 00.5

pH

mol

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A solution formed by mixing 50.0 mL of 0.020 M HCl with 125 mL of 0.010 M HI.

pH=?

H mol 00125.0soln L 1

mol 010.0

mL 1000

soln L 1mL 125

H mol 001.0soln L 1

mol 020.0

mL 1000

soln L 1mL 0.50

89.1)0129.0log(

H 0129.0L 175.0

H mol 00225.0

0.125L)(0.050L

H mol 00225.0

pH

M