A solution is made by mixing 500 ml of 0.172 M NaOh with exactly 500 ml of 0.100 M CH3COOH. Calculate the equilibrium concentration of the species below.
Ka of CH3COOH is 1.8 x10^-5
[H+]=?
{OH-]=?
[CH3COOH]=?
[Na+]=?
[CH3COO-]=?
mols NaOH = M x L = approx 0.0860
molw CH3COOH = approx 0.05 = HAc
........NaOH + HAc ==> NaAc + H2O
I....0.0860.....0.......0.......0
Add...........0.050..............
C.....-0.050.-0.050....0.050...0.050
E......0.036....0......0.050
So you have 0.0360 mols NaOH
0 mols HAc
0.050 mols NaAc = CH3COONa
all in 500 + 500 = 1000 mL H2O(assuming the volumes are additive).
M = mols/L each except OH^-
You find that from the hydrolysis of the Ac^- which is done as follows:
(Ac^-) = 0.05mols/L = 0.05
.........Ac^- + HOH ==> HAc + OH^-
I........0.05.............0....0
C.........-x..............x.....x
E.......0.05-x............x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.05-x) and solve for x = (OH^-). From that you can obtain H^+.
Post your work if you get stuck.
To find the equilibrium concentrations of the species in the solution, we can make use of the information given and apply the principles of acid-base equilibrium. Let's start by setting up the balanced equation for the reaction between CH3COOH (acetic acid) and NaOH (sodium hydroxide):
CH3COOH + NaOH ⇌ CH3COONa + H2O
The first step is to determine the initial moles of acetic acid (CH3COOH) and sodium hydroxide (NaOH) in the solution. To do this, we'll use the given molarity (M) values and the formula:
moles = volume (L) x molarity
For acetic acid (CH3COOH):
moles CH3COOH = 0.500 L x 0.100 mol/L = 0.050 mol
For sodium hydroxide (NaOH):
moles NaOH = 0.500 L x 0.172 mol/L = 0.086 mol
Since acetic acid and sodium hydroxide react in a 1:1 ratio, the moles of both species will be completely consumed when they reach equilibrium.
Next, we need to determine the change in moles for CH3COOH and NaOH at equilibrium. Since the reaction is in a 1:1 ratio, the change will be equal to the initial moles:
Δmoles CH3COOH = -0.050 mol
Δmoles NaOH = -0.086 mol
Now, let's calculate the equilibrium moles for each species:
Moles of CH3COOH at equilibrium: moles CH3COOH = initial moles CH3COOH + Δmoles CH3COOH
= 0.050 mol + (-0.050 mol) = 0 mol
Moles of NaOH at equilibrium: moles NaOH = initial moles NaOH + Δmoles NaOH
= 0.086 mol + (-0.086 mol) = 0 mol
Since both CH3COOH and NaOH have reacted fully, they are completely consumed and no longer present in the solution. This means their equilibrium concentrations are zero:
[CH3COOH] = 0 M
[Na+] = 0 M
To calculate the equilibrium concentrations of H+ and OH-, we need to consider the fact that water (H2O) undergoes an autoprotolysis reaction:
H2O ⇌ H+ + OH-
Water molecules can act as both acids and bases. In this case, since acetic acid (CH3COOH) is a weak acid and sodium hydroxide (NaOH) is a strong base, the OH- ions from NaOH will react with water to form more water, while the H+ ions from CH3COOH will donate protons to water and increase the concentration of H+ ions.
The key point to note here is that the concentration of H+ and OH- ions will change, but their product, which is the self-ionization constant (Kw) of water, will remain constant:
Kw = [H+][OH-] = 1.0 x 10^(-14) at 25°C
Since we know the value of Kw and the concentration of one ion (H+ or OH-), we can calculate the concentration of the other ion.
[H+] = Kw / [OH-]
Given that the Ka of acetic acid (CH3COOH) is 1.8 x10^-5, we can determine the concentration of H+ ions.
Ka = [H+][CH3COO-] / [CH3COOH]
Since the moles of CH3COOH and CH3COO- are equal (1:1 ratio), the concentration of CH3COOH at equilibrium will be the same as the concentration of CH3COO-:
[CH3COO-] = [CH3COOH]
Substituting these values into the expression for Ka:
1.8 x10^-5 = [H+][CH3COO-] / [CH3COOH]
Since [CH3COO-] = [CH3COOH], we can substitute [CH3COO-] with [CH3COOH]:
1.8 x10^-5 = [H+][CH3COOH] / [CH3COOH]
1.8 x10^-5 = [H+]
Therefore, [H+] = 1.8 x10^-5 M.
To determine [OH-], we can substitute this value into the expression for Kw:
Kw = [H+][OH-]
1.0 x 10^(-14) = (1.8 x10^-5)([OH-])
[OH-] = (1.0 x 10^(-14)) / (1.8 x10^-5)
Calculating this gives [OH-] = 5.6 x 10^(-10) M.
Finally, to summarize the equilibrium concentrations:
[H+] = 1.8 x10^-5 M
[OH-] = 5.6 x 10^(-10) M
[CH3COOH] = 0 M
[Na+] = 0 M
[CH3COO-] = 0 M