1. Calculate the value of the equilibrium constant for each of the following rea
ID: 826232 • Letter: 1
Question
1. Calculate the value of the equilibrium constant for each of the following reactions in aqueous solution:
a. HC2H3O2 + OH-? C2H3O2- + H2O
b. C2H3O2- + H+? HC2H3O2
c. HC2H3O2 + NH3? C2H3O2- + NH4+
2. Calculate the pH at the equivalence point for each of the following titrations:
a. 0.104 g of sodium acetate (Kb= 5.6 x 10-10) id dissolved in 25 mL of water and titrated with 0.0996 M HCl
b. 50.00 mL of 0.0426 M HOCl (Ka = 3.5 x 10-8) is titrated with 0.1028 M NaOH
c. 50.0 mL of 0.205 M HBr is titrated with 0.356 M KOH
Explanation / Answer
1)
Initially, let's discuss how the equilibrium constant, "K", is derived and thus why "K" represents the ratio of the "forward rate constant" divided by the "reverse rate constant".
For an unimolecular elementary step (a reaction that contains one reactant) that's considered the "slow" or "rate determining" step in a one step mechanism:
A ?? B (slow)
**Since the "Rate law" can be derived from experimental results organized in a chart OR from a "slow" elementary step, the rate law (which involves JUST the "reactants" of a specific equation) for the:
a) Forward reaction: With the forward equation being, A ? B
Rate forward = kforward [A]^1
b) Reverse reaction: With the reverse equation being, B ? A
Rate reverse = kreverse [B]^1
And since both equations are EQUAL to one another, we can set
Rate forward = Rate reverse
Thus, we can replace what the terms stand for:
kforward [A]^1 = kreverse [B]^1
Thus, after rearranging the equation:
kforward / kreverse = [B]^1 / [A]^1
THUS, since "kforward / kreverse" represents the equilibrium constant "K"...
K = [B]^1 / [A]^1
Thus, the equilibrium expression involves the concentration of the product of the forward reaction, "B", divided by the concentration of the reactants of the forward reaction, "A".
**When writing your equilibrium constant, make sure there are NO "solids", "pure liquids", or "inert gases".
**And since rate constants, "k", have their VALUES ONLY affected by TEMPERATURE, the VALUE of the equilibrium constant, "K", (which represents a ratio of two rate constants) is ALSO affected by TEMPERATURE.
For "a")
Since NaC2H302(s) is an ionic compound (thus electrolyte or salt) that is (due to solubility rules) completely SOLUBLE/miscible in the solvent water, it will dissociate completely into ions. Thus, the equilibrium constant, K, will have almost NO reactants (NaC2H302(s)) because almost all the products (Na+(aq) & C2H302-(aq)) will have formed because of it's ionic solubility. Since there will be a small denominator while there's a large numerator, thus the K will be a large value.
After dissociating, C2H3O2-(aq) will be able to act as a Bronsted-Lowry (and thus Lewis) base and react with H2O(l) to form OH-(aq) ions and the conjugate acid (of C2H3O2-) HC2H3O2(aq).
C2H302-(aq) + H2O(l) ?? OH-(aq) + HC2H3O2(aq)
The equilibrium expression will be:
Kb of C2H3O2-(aq) = [OH-(aq)]^1 [HC2H3O2(aq)]^1 / [C2H302-(aq)]^1
**EXCLUDING the pure liquid H2O.
For "b")
Since acetic acid is a WEAK "Bronsted-Lowry" (and thus "Lewis) acid, most of the acetic acid molecules will NOT dissociate much of their protons (H+(aq)) in the presence of water (to form H3O+(aq)) and thus there will not be many products of the forward reaction. And since products are the numerator in the equilibrium expression, the equilibrium constant "K" will be predicted to have a small value.
HC2H302(aq) + H2O(l) ?? H3O+(aq) + C2H3O2-(aq)
K = [H3O+(aq)]^1 [C2H3O2-(aq)]^1 / [HC2H302(aq)]^1
**Specifically, "K" should be "Ka" (the "acid dissociation" equilibrium constant") because an acid is involved in the reaction. For bases (like C2H3O2-(aq)), the "K" would specifically be "Kb". For other non acidic or basic reactions, if the species in the reaction are measured in terms of "concentration", then the "K" is specifically "Kc", while if the species are measured in pressure, the "K" is specifically "Kp".
**Notice how I excluded the PURE liquid "H2O" from the expression, because it's concentration will not change.
For "c") The relationship for an acid and it's conjugate base (or vice versa) is explained in the equation:
Kw = Ka x Kb
Where "Kw" is the equilibrium constant for water, "Ka" is the equilibrium constant for the acid, & "Kb" is the equilibrium constant for the (conjugate) base.
Thus, to find the Kb C2H3O2-(aq), we can divide Kw with the given Ka of HC2H3O2(aq):
Kb C2H3O2-(aq) = Kw / Ka of HC2H3O2(aq)
Kb C2H3O2-(aq) = 1E-14 / 5.81E-6
Kb C2H3O2-(aq) = 1.72E-9
There is a reason for why this equation explains the conjugate acid/base pair relationship, and it applies Hess's Law.
Example: For the conjugate acid/base pair of NH4+(aq) and NH3(aq)...
NH4+(aq) + H2O(l) ?? H3O+(aq) + NH3(aq) and the Ka of NH4+(aq)
The weak amine base NH3(aq) formed from the above reaction will react with H2O:
NH3(aq) + H2O(l) ?? OH-(aq) + NH4+(aq) and the Kb of NH3(aq)
Now if we add up to two equations via Hess's Law (and thus cross out similar species), we get the overall equation of:
H2O(l) + H2O(l) ?? H3O+(aq) + OH-(aq) and "new K"
**Thus, when equations are added together via "Hess's Law", the equilibrium constants are MULTIPLIED.
Thus:
"new K" = Ka of NH4+(aq) x Kb of NH3(aq)
**And this NET (overall) ionic equation is the "AUTO-IONIZATION" equation of water.
"new K" = Kw (which is 1E-14 @ 25?C or 298 Kelvin)
THUS, Kw = Ka of NH4+(aq) x Kb of NH3(aq)
2) Moles HClO = 0.0500 L x 0.0426 =0.00213
moles OH- required to reach the equivalnce point = 0.00213
Volume NaOH = 0.00213 / 0.1028 M=0.0207 L
total volume = 0.0207 + 0.0500= 0.0707 L
HClO + OH- = ClO- + H2O
moles ClO- = 0.00213
[ClO-]= 0.00213 / 0.0707 L=0.0301 M
ClO- + H2O <-------> HClO + OH-
K b= Kw/Ka = 1.0 x 10^-14 / 3.5 x 10^-8 =2.9 x 10^-7 = x^2/ 0.0301-x
x = [OH-]= 9.3 x 10^-5 M
pOH = 4.0
pH = 14 - 4.0 = 10
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