Question

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

Let us start by writing out the equations one by one.

H₂ + Cl₂ ----à 2HCl

We assume ideal gas behaviour to obtain the moles of HCl produced. We have the following parameters supplied.

P = 6.52 atm

V = 10.0 L

T = 300 K

Using the ideal gas law, we have (6.52 atm)(10.0 L) = nR(300 K)

w here n = amount of the ideal gas in moles and R = universal gas constant having a value of 0.082 L-atm/mol.K

Putting the values, we obtain n = (6.52 atm)(10.0 L)/(0.082 L-atm/mol.K)(300 K)

or, n = 2.6504 mole

Now, 2.6504 mole HCl is dissolved in 1.0 L water, so the molarity of the resulting solution is (2.6504 mol)/(1 L) = 2.6504 M

Now, we move to the other part of the problem. Let the original solution contain x mole of acetic acid and y mole of sodium acetate buffer so that the molarity of the solution is (x+y)/2.50 M. Now, we add 107 mL of HCl solution to 405 mL of the buffer solution and the pH is found to be 4.35.

Acetic acid is a weak acid and its dissociation in water follows:

CH₃COOH (aq) + H₂O (l) ------à H₃O⁺ (aq) + CH₃COO^- (aq)

K_a = [H₃O⁺][CH₃COO^-]/[CH₃COOH] = (z)(y/2.50)/(x/2.50) = (z)(y)/(x) where z is the initial mole of H₃O⁺ produced by dissociation of the weak acid.

Amount of HCl added = (107 mL/1000 L)(1 L)(2.6504 mol/L)= 0.2836 mole.

This added HCl will protonate sodium acetate to produce acetic acid. Now, 0.2836 mole of proton will protonate exactly 0.2836 mole of sodium acetate and produce 0.2836 mole of acetic acid as per the following reaction:

CH₃COO^- + H⁺ --------à CH₃COOH

and acetic acid will dissociate as per the above equation. Now, we must first find out the amounts of CH₃COOH and CH₃COO^- before addition of HCl.

Amount of CH₃COOH in 405 mL solution = (405 mL/1000 mL)(1L)(x/2.50 mol/L) = 0.162x mol

Similarly, amount of acetate in the solution = (405 mL/1000 mL)(1L)(y/2.50 mol/L) = 0162y mol.

After the reaction, the amounts of both CH₃COOH and CH₃COO^- have changed. Amount of CH₃COOH will be (0.162x + 0.2836) mole and that of acetate will be (0.162y – 0.2836) mole. The volume has changed. The new volume is now (107 + 405) = 512 mL.

So, the molar concentrations of CH₃COOH and CH₃COO^- after the addition of HCl will be

{(0.162x + 0.02836)mole/512 mL)(1000 mL/1 L) = 1000(0.162x + 0.2836)/512 M and 1000(0162y – 0.2836)/512 mol.

Also, the pH of the solution is 4.35. Therefore, 4.35 = - log₁₀[H₃O⁺]

or, [H₃O⁺] = 10^-pH = 10^-4.35 = 4.4668 x 10^-5

Now, finally, we add phenolpthalin to the solution and titrate with NaOH solution till the end point is reached.

We write the reaction between CH₃COOH and NaOH as

CH₃COOH + NaOH ------à CH₃COO^-Na⁺ + H₂O.

The reaction takes place on a 1:1 molar ratio. The amount of NaOH added is (657 mL/1000 mL)(1L)(0.920 mole/L) = 0.6044 mole.

The acid solution definitely contained 0.6044 mole acetic acid (since the two reacts on a 1:1 molar ratio). Now, we have already showed before that the acid solution contained (0.162x + 0.2836) mole acetic acid (we calculated that in the last step). Hence,

0.162x + 0.2836 = 0.6044

or, 0.162x = 0.3208

or, x = 1.9802

Our final calculation will entail the calculation of y. Since we have the acetic acid system, we can use its K_a value (1.75 x 10^-5) (we must assume that all the reactions are taking place at room temperature, 298 K). Therefore, pK_a = -log₁₀(1.75 x 10^-5) = 4.75

Now, we know from Henderson’s equation,

pH = pK_a + log₁₀[CH₃COO^-]/[CH₃COOH]

or, 4.35 = 4.75 + log₁₀[CH₃COO^-]/[CH₃COOH]

or, (-0.4) = log₁₀[CH₃COO^-]/[CH₃COOH]

or, [CH₃COO^-]/[CH₃COOH] = 10^(-0.4) = 0.398

Here, we must mention that the concentrations are molar concentrations and we have already calculated them in terms of x and y in the previous step.

Therefore, [1000(0.162y – 02836)/512]/[1000(0.162x + 0.2836)/512] = 0.398

or, (0.162y – 0.2836)/(0.162x + 0.2836) = 0.398

or, (0.162y – 0.2836) = 0.398(.0162x + 0.2836)

Putting the value of x that we have calculated (x = 1.9802), we have,

(0.162y – 0.2836) = 0.398(0.162*1.9802 + 0.2836) = 0.398(0.6044) = 0.2405

or, 0.162y = 0.5241

or, y = 3.2355

Molarity of acetic acid in the original solution = (1.9802 mole)/(2.50 L) = 0.792 M and that of sodium acetate = (3.2355 mole)/(2.50 L) = 1.294 M (ans)