calculate the ph of resulting solution when 2.50 ml of the 2.60 M acetic acid is diluted to make a 250.0 ml solution
Let's call acetic acid, CH3COOH, HAc. It saves time in typing.
(HAc) = 2.60M x (2.5/250) = 2.60 x (1/100) = 0.0260 M
Next set up an ICE table.
................HAc ==> H^+ + Ac^-
begin.........0.0260....0......0
change..........-x......+x....+x
final.......0.0260-x.....x......x
Ka = (H^+)(Ac^-)/(HAc)
Substitute from the ICE table into Ka expression and solve for x, then convert to pH.
Oh! wait but don't i have to plug it into ka=x^2/(0.0260-x) , ka of the acetic acid is 1.8 x 10^-5 and then solve for x ?
To calculate the pH of the resulting solution, you need to consider the dilution and the dissociation of acetic acid.
Step 1: Calculate the number of moles of acetic acid used.
Number of moles of acetic acid = volume (in L) × concentration (in mol/L)
= 2.50 mL × (1L/1000mL) × 2.60 mol/L
= 0.00650 mol
Step 2: Determine the concentration of acetic acid after dilution.
Final concentration of acetic acid = initial moles / final volume
= 0.00650 mol / 250 mL
= 0.000026 mol/mL
Step 3: Calculate the pOH of the solution.
pOH = -log10 (hydroxide ion concentration)
= -log10 ([OH^-])
= -log10 (Kw / [H3O+])
= -log10 (10^-14 / [H3O+])
= -log10 (10^14 × [H3O+])
= -(-14) - log10 ([H3O+])
= 14 - log10 ([H3O+])
Step 4: Calculate the pH of the solution.
pH + pOH = 14
pH = 14 - pOH
Step 5: Substitute the pOH value to calculate the pH.
pOH = -log10 ([OH^-])
= -log10 (0.000026 mol/mL)
= -log10 (2.6 × 10^-5)
= -(-4.58)
= 4.58
pH = 14 - 4.58
= 9.42
Therefore, the pH of the resulting solution is approximately 9.42.
To calculate the pH of the resulting solution, we need to determine the concentration of the diluted solution first. We'll follow these steps:
Step 1: Calculate the number of moles of acetic acid used.
molarity (M) = moles (mol) / volume (L)
Rearranging the equation, we get:
moles (mol) = molarity (M) × volume (L)
Given:
Molarity of acetic acid solution (initial) = 2.60 M
Volume of acetic acid solution used = 2.50 mL = 0.00250 L
Using the formula:
moles (mol) = 2.60 M × 0.00250 L
Step 2: Calculate the new concentration after dilution.
To find the concentration after dilution, we'll use the equation:
initial concentration × initial volume = final concentration × final volume
Given:
Initial concentration = 2.60 M
Initial volume = 2.50 mL = 0.00250 L
Final volume = 250.0 mL = 0.250 L
Using the equation:
2.60 M × 0.00250 L = final concentration × 0.250 L
Solving for the final concentration:
final concentration = (2.60 M × 0.00250 L) / 0.250 L
Step 3: Calculate the pH of the resulting solution.
The pH of a solution can be determined using the formula:
pH = -log[H+]
Since acetic acid is a weak acid, it partially dissociates in water, so we'll need to consider the equilibrium expression.
The dissociation equation for acetic acid (CH3COOH) is:
CH3COOH ⇌ CH3COO- + H+
Given that the concentration of acetic acid becomes the concentration of acetoacetate (CH3COO-) after dissociation, we can use the final concentration of acetic acid from step 2 as the concentration of acetoacetate (CH3COO-).
We'll assume that the equilibrium concentration of acetic acid (CH3COOH) is very close to the initial concentration.
Using the definition of pH:
pH = -log[H+]
Calculating the pH:
[H+] = acetoacetate concentration
pH = -log(acetoacetate concentration)
Enter the final concentration of acetoacetate (CH3COO-) from step 2 into the pH calculation.
Remember to keep the significant figures consistent throughout the calculation.
This will give you the pH of the resulting solution.