Question

4- Hydrogen and chlorine gas are placed in a 10.0 L container and allowed to react. The initial density of the mixture is 5.01 x10-3 gm/ml. The vessel is at 300K and the initial pressure is 6.52 atm. After the reaction is complete , the pressure is 6.52 atm. The gas is bubbled into 1.00 liter of deionized water. In another part of the universe, a grey haired chemistry instructor prepares an acetic acid/ acetate buffer. The buffer was prepared by adding acetic acid and sodium acetate to 2.50 L of deionized water. Now, as universes collide, our hero adds 107 ml of the HCl solution to 405 ml of the buffer. The pH is measured and is 4.35 after the HCl addition. A drop of phenolphthalein is added and the solution is titrated to equivalence with 657 ml of 0.920 M NaOH. Calculate the original molarities of the weak acid and conjugate base in the 2.50 L solution?

Answer 1

Solution.

The concentration of HCl solution is

c = n/V, where n is the number of moles.

As the initial density is 0.00501 g/mL, the total mass of gases is 0.00501g/mL*10000mL = 50.1 g.

The total number of gas moles is (n = PV/(RT)):

Converting pressure to SI units.
P = 660639 Pa
Converting volume to SI units.
V = 0.01 m³
T = 300 K
R = 8.314 J/(mol*K)
Computing number of moles using the selected equation.
n = 2.6486 moles;

The average molar mass of a gas mixture is M = m/n = 50.1g/2.6486mol = 18.92 g/mol.

On the other hand, the average molar mass of a gas mixture is

It means that Cl₂ is the limiting reactant, and the amount of Cl₂ is 0.245*2.6486= 0.649 moles.

As the reaction H₂+Cl₂ = 2HCl, the amount of HCl is 2*0.649 =1.298 moles.

Therefore, the concentration of HCl is 1.298 mol/1.00L = 1.298 M.

The pH of acetate buffer is determined by the molarities of the weak acid and conjugate base according to the equation

pK_a = 4.75 (reference data);

The concentration of HCl solution after the addition of the buffer solution is

After the addition of HCl the concentration of HAc became greater by 0.271 M and the concentration of NaAc became less by 0.271 M,

The titration with NaOH first recovered HAc formed by the addition of HCl, and 0.920*0.657-1.298*0.107 = 0.466 moles of NaOH were used to titrate the "initial" conjugate acid. Therefore, the amount of "initial" HAc is the same, namely, 0.466 moles, and its concentration in the initial buffer solution is 0.466/2.5 = 0.186 M. The concentration due to a dilution is 0.186*0.405/(0.107+0.405) = 0.147 M.

Now the equation is of only one variable:

[NaAc] = 0.437 M, or in the initial buffer solution, 0.437*(0.107+0.405)/0.405 =0.552 M.

Answer: [HAc] =0.186 M; [NaAc] = 0.552 M.