2. Consider the acid-dissociation reaction of nitrous acid (HNO2) a. Write the b

Question

2. Consider the acid-dissociation reaction of nitrous acid (HNO2) a. Write the balanced chemical equation for this reaction Make sure to write the plysical state of each species (10 points) b. Write the acidity constant expression for this reaction. Note that you do not have to determine the numerical value. (10 points) 3. Consider the HF solution with the concentration of 50% (w/v). You can convert the % concentration into the molarity. Answer the following questions. a. Assume that you have 100 mL. (exact) of this solution, and determine the number of moles of HF in 100 mL. (exact) of this solution. (10 points) b. Determine the molarity of HF by using the moles of HF from Question 3-a and the volume of the solution that you assumed in Question 3-a. (10 points)

Explanation / Answer

2a

balanced reaction

HNO2(aq) + H2O(l) <-> H3O+(aq) + NO2-(aq)

b)

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Ka = [H+][NO2-]/[HNO2]

also stated as

Ka = [H3O+][NO2-]/[HNO2]

Q3

a and b

m = 5 g per 100 mL so

5 g of HF

mol = mass/MW = 5/20 = 0.25 mol

V= 0.1 L

[HF] = 0.25/0.1 = 2.5 M

Q4

Ka = [CH3COO-][H+]/[CH3COOH]

Ka = (0.004)(0.004)/(0.8537)

Ka = 1.87*10^-5

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