1. You make a 2.00L solution of sodium acetate by adding 45.6 g of solid sodium

Question

1. You make a 2.00L solution of sodium acetate by adding 45.6 g of solid sodium acetate to water. What is the pH of the solution?
2. Dinitrogen tetroxide partially decomposes according to the following equilibrium: N2O4(g) 2NO2(g) A 2.000-L flask is charged with 8.28 × 10-3 mol of N2O4. At equilibrium, 5.14 × 10-3 mol of N2O4 remains. What is the Kc for this reaction?
3. Determine the equilibrium concentration of CO2(g). NH2COONH4(s) 2 NH3(g) + CO2(g) Kc = 1.58 × 10-8

1. You make a 2.00L solution of sodium acetate by adding 45.6 g of solid sodium acetate to water. What is the pH of the solution?
2. Dinitrogen tetroxide partially decomposes according to the following equilibrium: N2O4(g) 2NO2(g) A 2.000-L flask is charged with 8.28 × 10-3 mol of N2O4. At equilibrium, 5.14 × 10-3 mol of N2O4 remains. What is the Kc for this reaction?
3. Determine the equilibrium concentration of CO2(g). NH2COONH4(s) 2 NH3(g) + CO2(g) Kc = 1.58 × 10-8

1. You make a 2.00L solution of sodium acetate by adding 45.6 g of solid sodium acetate to water. What is the pH of the solution?
2. Dinitrogen tetroxide partially decomposes according to the following equilibrium: N2O4(g) 2NO2(g) A 2.000-L flask is charged with 8.28 × 10-3 mol of N2O4. At equilibrium, 5.14 × 10-3 mol of N2O4 remains. What is the Kc for this reaction?
3. Determine the equilibrium concentration of CO2(g). NH2COONH4(s) 2 NH3(g) + CO2(g) Kc = 1.58 × 10-8

Explanation / Answer

1) molarityof CH3COONa = (W/M)*(1/v)

       W = weight of CH3COONa = 45.6g

    M = molarmass of CH3COONa = 82 g/mol

   V = vol of solution = 2 L

Molarity(M) = (45.6/82)*(1/2) = 0.278 M

pH of CH3COONa = 7+1/2(pka+logC)

    pka of ch3cooh = 4.74

   C= concentration of ch3coona = 0.278 M

pH = 7+1/2(4.74+log0.278)

     = 9.092

2)          N2O4(g) <=====> 2NO2(g)

initial 8.23*10^-3/2           0

       4.115*10^-3       

change    1.545*10^-3 M         3.09*10^-3 M

equil       2.57*10^-3 M        3.09*10^-3 M

   Kc = [NO2]^2/[N2O4]

      = (3.09*10^-3)^2/(2.57*10^-3)

      = 0.0037

3)   NH2COONH4(s) 2 NH3(g) + CO2(g)

Kc = 1.58 × 10-8

Kc = [NH3]^2[CO2]

1.58*10^-8 = (2X)^2*X

X = 1.58*10^-3

AT equilibrium,

[CO2] = X = 1.58*10^-3 M

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