calculate the ph 25.00 mL solution of 0.125 M acetic acid (Ka= 1.75 x 10^-5) after the addition of each of the following volumes of 0.100 M KOH: 10.00 mL, Vep and 35.00 mL
KOH + HAc ==> H2O + KAc
Divide the problem up into
a. beginning (I've worked this but the problem doesn't ask for it.)
b. equivalence point
c. everything between a and b
d. everything after b.
What volume is b?
mols HAc = 0.025 x 0.125 = 0.003125
mols KOH = 0.003125(The coefficients are 1 mol KOH to 1 mol HAc.)
M KOH = mols KOH/L KOH solve for L; I obtained 31.25 mL. The pH is determined by the hydrolysis of the salt. Salt concn is M = mol/L = 0.003125/(25.00+ 31.25) = about 0.055M (that's an estimate)
.......Ac^- + HOH ==> HAc + OH^-
I....0.055.............0.....0
C.......-x.............x......x
E.....0.055-x .........x......x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.055-x)
Solve for x = OH^- and convert to pH.
a........ HAc ==> H^+ + Ac^-
I.......0.125.....0......0
C..........-x.....x......x
E......0.125-x....x......x
Substitute into Ka expression and solve for x = (H^+) and convert pH.
c. Use the Henderson Hasselbalch equation for these.
d. mols HAc = ?
mols KOH = ? and this is greater than mols HAc.
Excess KOH = subtract mols KOH-mols HAc.
Excess = KOH = OH^- ? and convert to pH.
To calculate the pH of a solution after the addition of a base, you need to consider the reaction between the acid and the base. In this case, acetic acid (CH3COOH) is a weak acid, and potassium hydroxide (KOH) is a strong base. The reaction between them will result in the formation of water (H2O) and the acetate ion (CH3COO-).
To solve this problem, we will consider three scenarios: after the addition of 10.00 mL of KOH, after the addition of the equivalence point volume (Vep), and after the addition of 35.00 mL of KOH.
Step 1: Determine the moles of acetic acid:
n(CH3COOH) = C(CH3COOH) × V(CH3COOH)
= 0.125 M × 25.00 mL
= 0.003125 moles
Step 2: Calculate the moles of hydroxide ions (OH-) based on the reaction:
1 mole of acetic acid reacts with 1 mole of hydroxide ions.
Step 3: After adding 10.00 mL of 0.100 M KOH:
n(OH-) = C(KOH) × V(KOH)
= 0.100 M × 10.00 mL
= 0.001 moles
The excess OH- concentration = (moles of OH-)/(total volume of solution)
= 0.001 moles / (25.00 mL + 10.00 mL)
Now, we use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([A-]/[HA])
Where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the acetate ion (CH3COO-), and [HA] is the concentration of the acetic acid (CH3COOH).
Step 4: Calculate the concentration of the acetate ion (CH3COO-) after the addition of 10.00 mL of KOH:
[A-] = moles of acetate ion / total volume of solution
Step 5: Calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Compute pKa using the given equilibrium constant (Ka):
pKa = -log(Ka)
Finally, substitute these values into the equation to determine the pH.
Follow the same steps for each volume of KOH (Vep and 35.00 mL) to calculate the pH after each addition.