Question

3. Measure the pH of the solutions described in the Data Table below. Make sure the electrode bulb is fully immersed before measuring. pH Data for Acetate Buffers mL of 0.60 M HC HOA mL of 0.60 M NaC2H302 Solution pH 30 2 30 3 30 10 4 10 30 5 20 20 4. Explain the pH for the five solutions. Consider the relative amounts of acid and base in each.

Answer 1

3

CH₃COOH + NaOH --------------> CH₃COONa + H₂O

pH is given by the equation

pH = pKa + log{[conjugate base]/[Acid]}

pH = pKa + log{[ CH₃COONa]/[ CH₃COOH]}

The acid dissociation constant, Ka for acetic acid = 1.8 x 10^-5

pKa = 4.744

(i) [CH₃COOH] = 0.030 L x 0.60 M = 0.018 mole

[ CH₃COONa] = 0 x 0.60 M = 0

CH₃COOH + H2O --------------> CH₃COO- + H₃O+

Initial                     0.018                                                     0              0

Change                 -x                                                            +x           +x

Equilibrium         0.018-x                                                 x              x

Ka = x2/(0.018-x)

1.8 x 10^-5= x²/0.018

x = [H+] = 5.692 * 10^-4

pH = -log(5.692 * 10^-4)

pH = 3.24

(ii) [CH₃COOH] = 0 x 0.60 M = 0

[ CH₃COONa] = 30 x 0.60 M = 0.018 mole

CH₃COONa + H2O --------------> CH₃COO- + NaOH

CH₃COO- + H2O = CH₃COOH + OH-

Kb=[CH3COOH][OH-]/[CH3COO-]

KaKb= 10-14

Kb= 10-14 / 1.8⋅10⁻⁵= 5.555 x 10⁻¹⁰

CH₃COONa + H2O --------------> CH₃COO- + NaOH

/Initial                   0.018                                                     0              0

Change                 -x                                                            +x           +x

Equilibrium         0.018-x                                                 x              x

Kb = 5.555 *10^-10 = x2/0.018

X = [OH-] = 0.000003162

pOH = -log(0.000003162) = 5.5

pH = 14-5.5

pH = 8.5

(iii) [CH₃COOH] = .030 x 0.60 M = 0.018 mol

[ CH₃COONa] = .010 x 0.60 M = 0.006 mole

pH = pKa + log{0.006/0.018}

pH = 4.744 + -0.4771

pH = 4.267

(iv) [CH₃COOH] = .010 x 0.60 M = 0.006 mol

[ CH₃COONa] = .030 x 0.60 M = 0.018 mole

pH = pKa + log{0.018/0.006}

pH = 4.744 + 0.47712

pH =5.2511

(v) [CH₃COOH] = .020 x 0.60 M = 0.012 mol

[ CH₃COONa] = .020 x 0.60 M = 0.012 mole

pH = pKa + log{0.012/0.012}

pH = 4.744 + 0

pH = 4.744

Solution	Volume of CH3COOH (ml)	Volume of CH3COONa (ml)	pH
1.	30	0	3.24
2.	0	30	8.50
3.	30	10	4.27
4	10	30	5.25
5.	20	20	4.74

4.            The pH increases as the concentration of sodium acetate increases in the solution, and when the concentration is same for both acetic acid and its conjugate base, the ph becomes equal to the pKa value of the acid.

3.24 is the pH of the 0.018 mol acetic acid whereas 8.5 is the pH of 0.018 mol of sodium acetate.

Except the solution 2 all the solutions are acidic since the pH values are less than 7. Solution 2 with pH=8.5 is basic (sodium acetate)