Question

1.) If 12.05 mL of a standard 0.1080 M KOH solution reacts with 52.65 mL of CH₃COOH solution, what is the molarity of the acid solution?

2.)Give the oxidation number of nitrogen in each of the following. (Type your answer in the format of −1.)

(a) N₂H₄


(b) HNO₂


(c) NH₄⁺


(d) HNO₃

3.) Give the oxidation number of sulfur in each of the following. (Type your answer in the format of −1.)

(a) SOCl₂


(b) BaSO₄


(c) SO₂


(d) Na₂S

Answer 1

1) If 12.05 mL of a standard 0.1080 M KOH solution reacts with 52.65 mL of CH₃COOH solution, what is the molarity of the acid solution?

millimoles of KOH = 12.05 x 0.108 = 1.3014

CH3COOH + KOH --------------> CH3COOK + H2O

millimoles of acid = millimoles of base

52.65 x M = 1.3014

Molarity = 0.02472 M

3)

(a) SOCl₂

oxidation state calculation :

x + (-2) + 2 (-1) = 0

x = +4

oxidation number of sulfur = +4

(b) BaSO₄

oxidation number = +6

(c) SO₂

oxidation number = +4

(d) Na₂S

oxidation number = -2