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Determination of the Dissociation Constant of a Weak Acid

By: Linh Chieu, Fez Lari, Kim Le, and Quinn Wade
by Quinn Wade on 14 November 2012

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Transcript of Determination of the Dissociation Constant of a Weak Acid

Concept Determination of the Dissociation
Constant of a Weak Acid Pre Lab Questions Data Trial 2 1. Define Bronsted-Lowry acids and bases.
A Bronsted-Lowry acid is a substance that acts as a proton donor, where as a Brownsted-Lowry base is a substance that acts as a proton acceptor. 2. Differentiate between the dissociation constant and equilibrium constant for the dissociation of a weak acid, HA, in aqueous solution.
The dissociation of a weak acid, HA, in aqueous solution is demonstrated by the following equation: HA + H2O (l) ⇋ H+ (aq) + A- (aq)
The equilibrium constant for this reaction is therefore: Keq = ([H+][A-])/([HA]), where [H+] and[ A-] are the product concentrations, HA is the reactant concentration, and water is assumed to be a pure liquid and thus not included in the equilibrium expression.
Similarly, the acid dissociation constant for this expression is: Ka = ([H+][A-])/([HA]) , where [H+] and [A-] are the ion concentrations yielded by the original weak acid [HA]. It is a measure of the degree to which the weak acid dissociates into its respective ions at equilibrium.
From both expressions, it can be concluded that, in this specific case, the dissociation constant and equilibrium constant for the dissociation of a weak acid are the same. 3. Why isn’t the pH at the equivalence point always equal to 7 in a neutralization titration? When is it 7?
The pH at the equivalence point is not always equal to 7. In a strong base-weak acid neutralization titration, the weak acid yields only small amounts of protons, which are immediately neutralized by the base. Also, the strong conjugate base of the weak acid is also produced in the process, making the solution more basic. 4.What is the pKa of an acid whose Ka is 6.5 x 10-6?
pKa = - log Ka
= - log 6.5 x 10-6
= 5.2 5. Why must two electrodes be used to make an electrical measurement such as pH?
Two electrodes must be used to make an electrical measurement such as pH. One electrode is utilized to sense the H3O+ concentrations in the solution, while the other is used to develop a known potential that is to be used as a reference. 6. What is a buffer solution?
A buffer solution is a solution that undergoes an only slight change in pH upon the addition of a small of amount of acid or base. 7. The pH at one half the equivalence point in an acid-base titration was found to be 5.67. what is the value of Ka for this unknown acid?

pH = pKa = 5.67
- log pKa = 5.67
log pKa = - 5.67
Ka = antilog – 5.67 = 10- 5.67 = 2.14 x 10-6 8. If 30.15 mL of 0.0995 M NaOH is required to neutralize 0.302 g of an unknown acid, HA, what is the molecular weight of the unkwown acid?
NaOH (aq) + HA (aq) → H2O (l) + NaA (aq)
Moles of NaOH = M x V
= 0.0995 M x 30.15 mL x (1 L)/(1000 mL)
= 2.9999 x 10-3 mol NaOH
= 2.9999 x 10-3 mol HA

(2.9999 x 〖10〗^(-3) mol HA )/(0.302 g HA)=(1 mol HA)/(x )
X= (0.302 g HA )/(2.9999 x 〖10〗^(-3) mol HA) = 100.67 g/(mol HA) 9. If Ka is 1.85 x 10-5 for acetic acid, calculate the pH at one half of the equivalence point and at the equivalence point for a titration of 50 mL of 0.100 M acetic acid with 0.100 M NaOH. HC2H3O2 (aq ) + NaOH (aq) H2O (l) + NaC2H3O2HC2H3O2 (aq ) + H2O (l) H3O+ (aq) + CH3COO- (aq)At one half of the equivalence point:[HC2H3O2] = [CH3COO-] (50% of the acid has been titrated to produce CH3COO-) pH = pKa = - log Ka = - log 1.85 x 10-5 pH = 4.73At the equivalence point: Moles HC2H3O2 = moles NaC2H3O2 = 50 mL x (1 L)/(1000 mL ) x 0.100 M = 0.005 mol NaC2H3O2[NaC2H3O2] = (0.005 mol )/(0.05 L+0.05 L) = 0.05 MHydrolysis of CH3COO-: Ka = 1.8 x 10-5
CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq)
I 0.05 m 0 ≈ 0
C -x + x + x
E (0.05 – x) x x
Kb = ([CH3COOH][OH-])/([CH3COO]) = Kw/Ka = (1.0 x 10^-14)/(1.8 x 10^-5) = 5.6 x 10-10(x ×x)/(0.05-x) = 5.6 x 10-10

Assume x << 0.05,
so 0.05 – x = 0.05(x ×x)/0.05 = 5.6 x 10-10
X2 = 2.8 x 10-11
X = √(2.8 x 10^-11) = 5.3 x 10-6 = [OH-]

pH = 14.00 – pOH
= 14.00 – (-log [OH-])
= 14.00 – (-log 5.3 x 10-6)
= 14.00 – 5.28pH = 8.72 Trial 1:
mL pH
1 3.76
2 3.80
3 3.95
4 4.07
5 4.20
6 4.26
7 4.35
8 4.40
9 4.51
10 4.60
11 4.67
12 4.75
13 4.82
14 4.88
15 4.96
16 5.06
17 5.16 18 5.25
19 5.36
20 5.49
21 5.65
22 5.85
23 6.19
24 6.92
25 11.69
26 11.74
26.1 11.76
26.2 11.78
26.3 11.82
26.4 11.84
26.5 11.88
26.7 11.89
26.8 11.92
26.9 11.93
27 11.93 Trial 3 Practice Problems 1. If a solution has an [H+]concentration of 4.5 x 10-7, give pH, pOH, and whether it is acidic or basic. How acidic or basic??? 2. Explain why even a basic solution contains some H+ ions and why even an acidic solution contains some OH- ions. mL pH
1 3.50
2 3.62
3 3.77
4 3.92
5 4.08
6 4.19
7 4.27
8 4.38
9 4.44
10 4.51
11 4.59
12 4.68
13 4.70
14 4.80
15 4.88
16 4.95
17 5.04 Ka- is the acid ionization constant for equilibrium; Ka= [H+][A-]/[HA] Kb- is the base ionization constant for equilibrium; Kb=[BH+][OH-]/[B] Dissociation Constant- (Kd) a type of equilibrium constant that applies to the dissociation (separation) of a complex molecule into its subcomponents By:Quinn Wade, Linh Chieu, Kim Le, and Fez Lari 18 5.12
19 5.23
20 5.32
21 5.46
22 5.61
23 5.79
24 6.08
25 6.44
27 7.35
28 10.46
28.1 11.28
28.2 11.70
28.3 11.91
28.4 12.08
28.5 12.10
28.6 12.12
28.7 12.12 mL pH
1 3.27
2 3.57
3 3.76
4 3.94
5 4.06
6 4.17
7 4.27
8 4.37
9 4.45
10 4.53
11 4.61
12 4.68
13 4.76
14 4.86
15 4.94
16 5.00
17 5.10
18 5.21 Strength of Acids and Bases
19 5.34
20 5.44
21 5.59
22 5.79
23 6.05
24 6.78
25 6.88
26 11.24
26.1 11.37
26.2 11.46
26.3 11.48
26.4 11.51
26.5 11.53
26.6 11.59
26.7 11.59
26.8 11.62
26.9 11.66
27 11.70 VS The Dissociation Constant YUMMM!!!!! Strong Acid: an acid (hydrogen-containing compound that can produce H+ in aqueous solutions) that ionizes completely in water Strong Base: a base( produces excess of OH- whan it dissolves in water) that ionizes completely in water Weak Acid: an acid that only partly ionizes in water Weak Base: a base that only partly ionizes in water Ka >1 Ka <1 Kb >1 Kb <1 Ka= [H+][A-]/[HA] Kb=[BH+][OH-]/[B] product concentrations of H+ and A- much larger than concentration of HA due to extensive dissociation product concentrations of H+ and A- much smaller than concentration of HA due to slight dissociation product concentrations of BH+ and OH- much larger than concentration of B due to extensive dissociation product concentrations of BH+ and OH- much smaller than concentration of B due to slight dissociation Titration! Calculations D. Concentration of Unknown Acid
Volume of 0.09783 M NaOH pH
Trial 1 24.30 mL 8.5
Trial 2 27.20 mL 8.5
Trial 3 25.75 mL 8.5
Average 25.75 mL 8.5

Moles of NaOH:
25.75 mL x (1 L)/(1000 mL) x 0.09783 M NaOH = 2.5191 x 10-3 mol NaOH
Moles of unknown acid at the equivalence point:
moles of NaOH = 2.5191 x 10-3 mol weak acid
Molarity of Acid:
(mol acid)/(volume acid)=(2.5191 x 10-3 mol )/(25 mL x (1 L)/(1000 mL)) = 0.1008 M weak acid Logistic Equation Body Calcium Absorption f(x)= c/1+ae^-bx Titration: even out pH
cannot theoretically exceed limit: no further change
indicated balance Like titration, logistic equations have a set limit (horizontal asymptote/value c), which indicates the endpoint. The value remains constant no matter the continuation of the line. Like titration, calcium vitamin intake has a set limit. The body at one period of time can only absorb 600g of this nutrient; therefore, although there are 1200g pills the absorption is at its endpoint and remains constant no matter how much more is consumed. Ka= [H+][A-]/[HA] Kb=[BH+][OH-]/[B] D. Concentration of Unknown Acid
Molarity of NaOH:
Trial 1:
0.5000 g KHP x (1 mol KHP)/(204.23 g KHP) = 2.448 x 10-3 mol KHP ÷ 0.1 L = 2.448 x 10-2 M
MbVb = MaVa
Mb x 0.02538 L = 2.448 x 10-2 M x 0.1 L
Mb = 0.09646 M
Trial 2:
0.5020 g KHP x (1 mol KHP)/(204.23 g KHP) = 2.458 x 10-3 mol KHP ÷ 0.1 L = 2.458 x 10-2 M
MbVb = MaVa
Mb x 0.02485 L = 2.458 x 10-2 M x 0.1 L
Mb = 0.09891 M
Trial 3:
0.5020 g KHP x (1 mol KHP)/(204.23 g KHP) = 2.458 x 10-3 mol KHP ÷ 0.1 L = 2.458 x 10-2 M
MbVb = MaVa
Mb x 0.02505 L = 2.458 x 10-2 M x 0.1 L
Mb = 0.09812 M
Average molarity of NaOH:
(0.09646 M+0.09891 M+0.09812 M )/3 = 0.09783 M NaOH HA + NaOH -> H2O + NaA
NaA + H2O <-> H3O+ + OH-
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