I did an experiment using 20ml of 0.1M acetic acid and added 8ml of 0.1M NaOH. pH obtained after adding NaOH was 5.2. How do i calculate the mols of NaOH that reacted? mols of acetate formed, acetic acid initially present and acetic acid unreacted. Also how do i determine the [H+], [CH3CO2-] and [CH3CO2H] and Ka

To save some typing, I will call acetic acid, HAc and acetate is Ac^-

millimoles HAc = mL x M = 20 x 0.1 = 2.0
mmoles NaOH = 8 x 0.1 = 0.8

...............HAc + NaOH ==> NaAc + H2O
initially......2.0....0........0.......0
added.................0.8...............
change.........-0.8...-0.8.....+0.8..+0.8
equil.......... 1.2.....0......0.8...0.8

(HAc) = mmoles/mL = 1.2/28 = ??
NaAc = same process = mmoles/mL.

pH = pKa + log [(NaAc)/(HAc)] = ??

To calculate the moles of NaOH that reacted, you can use the concept of stoichiometry.

1. Start by determining the balanced chemical equation for the reaction between acetic acid (CH3CO2H) and sodium hydroxide (NaOH):

CH3CO2H + NaOH -> CH3CO2Na + H2O

2. Since the ratio of acetic acid to sodium hydroxide in the balanced equation is 1:1, the moles of NaOH reacted will be the same as the moles of acetic acid initially present.

Moles of NaOH reacted = Moles of acetic acid = 0.1 M * (20 mL / 1000 mL) = 0.002 moles

To determine the moles of acetate (CH3CO2-) formed, you need to consider the stoichiometry of the reaction. Since 1 mole of acetic acid reacts with 1 mole of sodium hydroxide, the moles of acetate formed will also be equal to 0.002 moles.

To find the moles of acetic acid initially present, you'll need to subtract the moles of acetic acid reacted from the total moles of acetic acid used.

Moles of acetic acid initially present = Moles of acetic acid - Moles of acetic acid reacted = 0.002 moles - 0.002 moles = 0 moles

Since all the acetic acid has reacted, there is no acetic acid left unreacted.

To determine the concentrations of [H+], [CH3CO2-], and [CH3CO2H], you can use the dissociation of acetic acid. Acetic acid is a weak acid that partially dissociates in water:

CH3CO2H (acetic acid) + H2O (water) ⇌ CH3CO2- (acetate) + H+ (hydrogen ion)

This equilibrium is described by the acid dissociation constant (Ka). The formula for Ka is:

Ka = [CH3CO2-] * [H+] / [CH3CO2H]

You know the pH, which is a measure of the H+ concentration. pH = -log[H+]. In this case, the pH is 5.2. To find the [H+], you need to convert the pH to a concentration:

[H+] = 10^(-pH) = 10^(-5.2) = 6.31 x 10^(-6) M

Now, you can rearrange the equation for Ka to solve for [CH3CO2-]:

Ka = [CH3CO2-] * [H+] / [CH3CO2H]

Rearranging the equation: [CH3CO2-] = (Ka * [CH3CO2H]) / [H+]

Plug in the values:

[CH3CO2-] = (Ka * 0.002 moles) / (6.31 x 10^(-6) M)

Finally, you can determine the concentration of acetic acid [CH3CO2H] using the equation:

[CH3CO2H] = 0.1 M - [CH3CO2-]

Substitute the value of [CH3CO2-] you found and calculate [CH3CO2H].