sorry,for the mistake.
To 200mL of a 0.1M solution of a CH3COOH (Ka=1.8 x 10-5) 70mL of a 0.2M solution of NaOH have been added.calculate the pH before and after the NaOH addition.
CH3COOH - acetic acid = HAc
CH3COONa = sodium acetate = NaAc
BEFORE: you have a 200 mL of 0.1M HAc. The pH is determined by the ionization of the HAc.
.........HAc ==> H^+ + Ac^-
I........0.1.....0......0
C.......-x.......x......x
E.......0.1-x....x......x
Substitute the E line into the Ka expression for HAc and solve for x = (H^+), then convert to pH.
AFTER.
millimols HAc = mL x M = 200 x 0.1 = 20
mmols NaOH added = 70 x 0.2 = 14
.....HAc + NaOH ==> NaAc + H2O
I....20.....0........0......0
add.........14...............
C...-14.....-14......14
E....6.......0.......14
Substitute the E line into the Henderson-Hasselbalch equation and solve for pH. Post your work if you get stuck.
why did you take 0.1M? Why don't you multify with volume in the first step?
To calculate the pH before and after the NaOH addition, we need to understand the reaction that occurs when CH3COOH (acetic acid) reacts with NaOH (sodium hydroxide). This is a neutralization reaction in which acetic acid reacts with sodium hydroxide to form sodium acetate (CH3COONa) and water (H2O).
The balanced equation for this reaction can be written as:
CH3COOH + NaOH -> CH3COONa + H2O
Before the addition of NaOH:
1. Calculate the initial amount of moles of acetic acid (CH3COOH) in the 200 mL of 0.1 M solution.
Moles of CH3COOH = volume (in liters) × molarity
Moles of CH3COOH = 0.2 L × 0.1 M = 0.02 moles (n1)
2. Calculate the initial amount of moles of NaOH in the 70 mL of 0.2 M solution.
Moles of NaOH = volume (in liters) × molarity
Moles of NaOH = 0.07 L × 0.2 M = 0.014 moles (n2)
3. Determine which of the two reactants, CH3COOH or NaOH, is the limiting reactant. The limiting reactant is the one that gets fully consumed and determines the amount of product formed. In this case, NaOH is the limiting reactant since it is present in a lesser amount.
4. Calculate the amount of moles of CH3COOH left after the reaction with NaOH.
Moles of CH3COOH left = n1 - n2 = 0.02 moles - 0.014 moles = 0.006 moles (n3)
5. Determine the concentration of acetic acid left after the NaOH reaction.
Concentration of CH3COOH left = moles (n3) / volume (in liters)
Volume = 0.2 L + 0.07 L = 0.27 L
Concentration of CH3COOH left = 0.006 moles / 0.27 L ≈ 0.022 M
Next, we can calculate the pH of the resulting solution after the reaction of acetic acid with NaOH.
1. Calculate the concentration of sodium acetate (CH3COONa) formed.
The number of moles of CH3COONa formed is the same as the number of moles of NaOH used (0.014 moles). Dividing this by the total volume:
Concentration of CH3COONa = 0.014 moles / 0.27 L ≈ 0.052 M
2. Find the concentration of OH- ions in the solution after the reaction.
Since NaOH is a strong base, it completely dissociates in water, producing an equal concentration of OH- ions. So, the concentration of OH- is 0.052 M.
3. Calculate the pOH using the concentration of OH- ions.
pOH = -log10(OH- concentration)
pOH = -log10(0.052) ≈ 1.28
4. Calculate the pH using the pOH value.
pH + pOH = 14
pH + 1.28 = 14
pH = 14 - 1.28 ≈ 12.72
Therefore, the pH after the addition of NaOH is approximately 12.72.
Please note that this calculation assumes there are no other acid-base reactions occurring in the system and that the volumes are additive.