Equilibrium:

A 2.0 l bulb contains 6.00 mol of NO₂, 3.0 mol NO and 0.20 mol of O2 at equilibrium. What is the K_eq for 2NO +O₂ ßà 2NO₂ ?

K_eq = [NO₂]² / [NO]²[O₂]

[NO₂] = 6.0 mol / 2.0 L = 3.0 M

[O₂] = 0.20 mol /2.0 L = 0.10 M

[NO] = 3.0mol / 2.0 L = 1.5 M

K_eq = (3.0)² / (1.5)²(0.10) = 40

4.00 mol of NO₂ is introduced into a 2.00 L bulb. After a while equilibrium is attained according to the equation 2NO +O₂ ßà 2NO_{2 .} at equilibrium 0.500 mol of NO is found. What is the Keq value?

K_eq = [NO₂]² / [NO]²[O₂]

[NO₂]start = 4.00 mol / 2.00 L= 2.00M

[NO]equil – 0.500 mol / 2.00 L = 0.250 M

2NO +O₂ ßà 2NO₂

I        0          0              2.00

C       +0.25    +0.125   -0.25        

E         0.25        0.125  1.75

K_eq = (1.75)² / (0.250)²(0.125) = 392

Solubility Equilibrium:

What is the concentration of all the ions present in a saturated solution of Ag₂CO₃ having concentration of 1.2 x 10^-4

Ag₂CO₃ à 2Ag⁺ +CO₃^2-

1 mol         2 mol  1 mol

[CO₃^2-] = 1.2 x 10^-4

[Ag⁺] = 1.2 x 10^-4 mol Ag₂CO₃ / L * 2 mol Ag⁺ / 1 mol l Ag₂CO₃

         = 2.4 x10^-4 M

If 5.0 ml of 0.020 M C^l- is added to 15.0 ml of 0.012 M Br^-, what is the molarity of the Cl- and Br- ions in the mixture?

[substance]diluted = [substance]old * old volume / diluted volume

[Cl-]diluted = 0.020 M x 5.0 ml/ (5.0 + 15.0) ml = 0.0050M

[Br^-]diluted =  0.012 x 15.0 ml / (5.0 + 15.0) ml = 0.0090 M

A solution in equilibrium with a precipitate of BaF₂ contains 4.59 * 10^-2 M Ba²⁺ and 2.00 x 10^-3 M F^-. What is K_sp for BaF₂?

BaF_{2 (s)} ßà Ba2+ + 2F^-

Ksp = [Ba2+][F-]2

Ksp = (4.59 x10^-2) (2.00 x10-3)2 = 1.84 x 10^-7

What is the [Mg²⁺] in a saturated solution of Mg(OH)₂ ?

Mg(OH)_{2 (s)} ßà Mg²⁺ + 2OH^-

Ksp = [Mg²⁺][OH]²

        = (x)(2x)²

4x³ = 5.6 x 10^-12

x³ = 1.4 x 10^-12

x = 1.1 x 10^-4 M

Acid Base and Salt:

If pH = 9.355, what is pOH?

Since: pH + pOH = 14.000

Then: pOH = 14.000 – pH = 14.000 – 9.355 = 4.645

If pH = 6.330, what is [OH^-]?

pOH = 14.000 – pH = 14.000 – 6.330 = 7.670

[OH^-] = antilog(-7.670) = 2.14 x 10^-8

If 10.0mL of 0.100 M HCl is added to 90.0 mL of 0.100 M NaOH, what is the pH of the mixture?

[H₃O⁺]_ST = 0.100 M x 10.0mL/100.0mL = 0.0100M

[OH^-]_ST = 0.100 M x 90.0 mL/100.0mL = 0.0900 M

[OH^-]_XS = [OH^-]_ST - [H₃O⁺]_ST

            = 0.0900 -0.0100 = 0.0800 M

pOH = -log(0.0800) = 1.097

pH = 14.000 – pOH = 14.000 – 1.097 = 12.903

If Ka = 1.8 * 10-5 for CH₃COOH, what is the pH of a 0.500 M solution of CH₃COOH?

CH₃COOH + H₂O ßà H₃O⁺ + CH₃COO^-

I        0.500                          0               0

                            C        - X                           +X             +X

                            E      0.500-X                          X               X

K_a= [C H₃COO^-][H₃O⁺]/[C H₃COOH]

    =  X² /0.500 – X = 1.8 * 10^-5

X = 3.0 * 10^-3 M = [H₃O⁺]

Assume 0.500 – X  =  0.500

pH = 2.52