Question

Problem #C4:

A) Calculate the Molarity for: H⁺(at equilibrium) and HCl (after dilution occurs but before the reaction occurs).

B) Calculate the Molarity for: H⁺(at equilibrium) and NaOH (after dilution occurs but before the reaction occurs).

C. pH of Buffer Solutions:

1. Preparation of Buffer Solution: Weigh 3.3 g of solid sodium acetate trihydrate, NaC2H3O2 * H2O, into a clean 150 or 250 mL beaker. Using graduated cylinders, add 46 mL of deionized water and 4.0 mL of 6.0 M HC2H3O2. Mix thoroughly. Measure 19.0 mL of this into each of two clean beakers. Label as "A" and "B". Measure the pH of this buffer solution using the remaining 10 mL.

2. Addition of Acid or Base to Buffer: a. To the buffer in beaker "A", add 1.0 mL of 3.0 M HCl. Mix thoroughly and read the pH of the resulting mixture. b. To the buffer in beaker "B,", add 1.0 mL of

3.0 M NAOH. Mix thoroughly and read the pH of the resulting mixture. 3. pH of Deionized water: Obtain 19 mL of deionized water in each of two beakers. Read the pH of the deionized water. (NOTE: It will not be a pH of 7.00.)

4. Addition of Acid or Base to Deionized Water: a. To one beaker of deionized water, add 1.0 mL of 3.0 M HCl. Mix thoroughly and read the pH of the resulting mixture. b. To the second beaker of deionized water, add 1.0 mL of 3.0 M NAOH. Mix thoroughly and read the pH of the resulting mixture.

Answer 1

I need the pKa value for this buffer, do you have it?. I found this, but I don't know if it's the same of your so:

pKa of acetic acid = 4.75 (sometimes is reported as 4.74 or even 4.8, but I'll use 4.74)

With the innitial data, we can know the pH of the buffer and the reaction that is taking place. first let's calculate the moles (MW of NaC2H3O2*3H2O reported is 136.07 g/mol)

moles of NaC2H3O2 * 3 H2O = 3.3 g / 136.07 g/mol = 0.024 moles

moles of acid = 6 mol/L * 0.004 L = 0.024 moles

to get pH, let's use HH equation:

pH = pKa + log [S] / [A]

pH = 4.74 + log (0.024 / 0.024)

pH = 4.74

So pH = pKa of the acid.

Now, the moles of HCl are:

moles HCl = 3 * 0.001 = 0.003 moles

When we add HCl to this solution, we are increasing the concentration of H⁺ and lowering the concentration of the acetate ion of this buffer so:

CH₃COOH ------> H⁺ + CH₃COO^-

i. 0.024 0.003 0.024

e. 0.024+0.003 0.024 - 0.003

now the HH equation can be written as:

pH = pKa + log acetate/ac. acid

pH = 4.74 + log (0.024-0.003/0.024+0.003)

pH = 4.74 - 0.11 = 4.63

Concentration of H⁺ in equilibrium: [H⁺] = 10^-pH = 10^-4.63 = 2.34x10^-5 M

The total volume is 19 mL + 1 mL = 20 mL or 0.020 L

Concentration of HCl after dilution occurs: 0.003 moles / 0.020 = 0.15 M

For the part b) we do exactly the same thing but instead of substract to the acetate, we substract the moles to the acid, cause with base we are increasing concentration of the acetate ion, nd decreasing the acid (pH will rise) the moles of NaOH are the same of HCl, so concentration would be the same (as we are using the same total volume of 20 mL)

[NaOH] = [HCl] = 0.15 M

pH = 4.74 + log (0.024+0.003/0.024-0.003)

pH = 4.74 + 0.11 = 4.85

[H⁺] = 10^-4.85 = 1.41x10^-5 M

If you need something else, or something to be fixed or explained better, tell me in a comment and I'll help you out.

Hope this helps