0.100 moles of NaOH are combined with 0.125 moles acetic acid. The mixture is taken to a final volume of 100.00 ml. What is the expected pH of the solution? Use Ka for acetic acid of 1.76 x 10^-5
It makes it easier if you don't change screen names. Use the HH equation as in in the formic acid/sodium formate buffer
There is one difference. In the formic/formate buffer you have the components separated. In this one you must combine the acid and base, calculate how much acetate you have formed and how much acetic acid there is left to act as the cid.
To determine the expected pH of the solution, we need to calculate the concentration of acetate ions (CH3COO-) and hydrogen ions (H+) in the solution. Then we can use the equation for the dissociation of acetic acid to calculate the pH.
1. First, let's calculate the concentration of acetate ions (CH3COO-) in the solution. Since acetic acid is a weak acid, we can assume that it is completely dissociated.
The moles of acetic acid present in the solution = 0.125 moles
The volume of the solution = 100.00 mL = 0.100 L
Therefore, the concentration of acetate ions = moles/volume = 0.125 moles / 0.100 L = 1.25 M
2. Next, let's calculate the concentration of hydrogen ions (H+). Since acetic acid is weak, a fraction of it will remain undissociated. To calculate this fraction, we need to use the dissociation constant (Ka) of acetic acid.
The equation for the dissociation of acetic acid is:
CH3COOH ⇌ CH3COO- + H+
The initial concentration of acetic acid = 0.125 moles
The final concentration of acetate ions (CH3COO-) = 1.25 M (from step 1)
Applying the equation for the dissociation of acetic acid and the Ka value, we have:
Ka = [CH3COO-][H+]/[CH3COOH]
(1.76 x 10^-5) = (1.25 M)[H+]/(0.125 M)
[H+] = (1.76 x 10^-5)(0.125 M) / 1.25 M
[H+] = 1.76 x 10^-6 M
3. Finally, let's calculate the pH of the solution using the concentration of hydrogen ions.
The pH is calculated using the formula: pH = -log[H+]
pH = -log(1.76 x 10^-6)
pH ≈ 5.76
Therefore, the expected pH of the solution is approximately 5.76.
To find the expected pH of the solution, we need to calculate the concentration of the acetic acid (CH3COOH) and the concentration of the sodium acetate (CH3COONa) formed when NaOH reacts with acetic acid.
We can start by calculating the final concentration of acetic acid (CH3COOH) and sodium acetate (CH3COONa) in the solution.
Given:
- Moles of NaOH = 0.100
- Moles of acetic acid = 0.125
- Final volume = 100.00 ml
First, convert the final volume from milliliters to liters:
Final volume = 100.00 ml = 100.00/1000 = 0.100 L
Next, calculate the concentration of acetic acid (CH3COOH):
Concentration of acetic acid = Moles of acetic acid / Final volume
= 0.125 mol / 0.100 L
= 1.25 mol/L
Since NaOH reacts in a 1:1 ratio with acetic acid, the concentration of sodium acetate (CH3COONa) will also be 1.25 mol/L.
Now, we can use the equilibrium expression for acetic acid (CH3COOH) to find the concentration of hydroxide ions (OH-) and then calculate the pOH and pH of the solution.
The equilibrium expression for acetic acid is:
CH3COOH ⇌ CH3COO- + H+
The equilibrium constant, Ka, is given as 1.76 x 10^-5. Let's call the initial concentration of acetic acid [CH3COOH] and the concentration of hydrogen ions [H+].
Using an ICE table:
Initial: [CH3COOH] 1.25
Change: -x x
Final: [CH3COOH]-x x
Since [CH3COOH] decreases by x after reacting, we can approximate [CH3COOH]-x as approximately 1.25 (since x is much smaller than 1.25).
Now we can set up the expression for Ka:
Ka = [CH3COO-][H+] / [CH3COOH]
1.76 x 10^-5 = (x)(x) / (1.25 - x)
Since x is small compared to 1.25, we can approximate 1.25 - x as 1.25.
1.76 x 10^-5 = x^2 / 1.25
Rearranging the equation:
x^2 = 1.76 x 10^-5 * 1.25
x^2 = 2.20 x 10^-5
Taking the square root of both sides:
x = sqrt(2.20 x 10^-5)
x ≈ 0.004691
Now we have the concentration of hydrogen ions [H+], which is equal to the concentration of hydroxide ions [OH-] since NaOH neutralizes the acetic acid.
Therefore, [H+] ≈ [OH-] ≈ 0.004691 mol/L
To find the pOH (-log[OH-]) and pH (-log[H+]), we can use the following:
pOH = -log[OH-]
pH = 14 - pOH
pOH ≈ -log(0.004691)
pOH ≈ -(-2.327)
pOH ≈ 2.327
pH ≈ 14 - 2.327
pH ≈ 11.673
Therefore, the expected pH of the solution is approximately 11.673.