Ok, moving on from my last question, you have 100mL of 0.10M acetic acid with ka = 1.8x10(-5) and add 50mL (0.10M) NaOH.
100 mL HAc x 0.1M = 10 mmoles.
50 mL NaOH x 0.1M = 5 mmoles.
.........HAc + OH^- => Ac^- + H2O
begin....10....0........0.......0
add............5.0.................
change..-5.0...-5.0.....+5.0....+5.0
equil....5.0.....0.......5.0
Then I would use the Henderson-Hasselbalch equation to solve for pH.
Since (base) = (acid), then log base/acid = log 1 = 0 so pH = pKa or about 4.74
To determine what happens when 100mL of 0.10M acetic acid with a Ka value of 1.8x10^(-5) is mixed with 50mL of 0.10M NaOH, we need to consider the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH). This reaction is known as neutralization.
The equation for the reaction is:
CH3COOH + NaOH → CH3COONa + H2O
In this case, we have a 1:1 stoichiometric ratio between acetic acid and sodium hydroxide. This means that for every one mole of acetic acid that reacts, one mole of sodium acetate (CH3COONa) is produced.
To determine the concentration of the resulting sodium acetate solution, we can use the dilution formula:
C1V1 = C2V2
Where:
C1 = initial concentration of acetic acid
V1 = initial volume of acetic acid
C2 = final concentration of sodium acetate
V2 = final volume of the resulting solution
In this case, we initially have 100mL of 0.10M acetic acid and add 50mL of 0.10M NaOH. Since the volumes are being added together, the final volume of the resulting solution is 100mL + 50mL = 150mL.
Using the dilution formula, we can calculate the final concentration of sodium acetate (C2):
(0.10M)(100mL) = C2(150mL)
C2 = (0.10M)(100mL) / (150mL)
= 0.0667 M
Therefore, the concentration of the resulting sodium acetate solution is 0.0667M.