200.0 ml of 0.5000 M acetic acid is added to 200.0 ml of 0.5000 M NaOH. What is
the final pH? The Ka of acetic acid at 25 oC is 1.76 x 10-5.
n(NH3COOH)=CxV=0.5x200/1000=0.1mol
n(Naoh)=CxV=0.5x200/1000=0.1mol
I have no idear of what to do after finding the number of moles
After finding the number of moles of acetic acid (NH3COOH) and sodium hydroxide (NaOH), you can use the stoichiometry of the reaction between acetic acid and sodium hydroxide to calculate the concentration of the remaining reactants and the product.
The balanced equation for the reaction between acetic acid and sodium hydroxide is:
NH3COOH + NaOH â Na(NH3COO) + H2O
The stoichiometry of the reaction tells us that one mole of acetic acid reacts with one mole of sodium hydroxide to produce one mole of sodium acetate (Na(NH3COO)) and one mole of water (H2O).
Since the initial concentration of acetic acid and sodium hydroxide is the same (0.5000 M) and the stoichiometry is 1:1, after the reaction is complete, the concentration of acetic acid and sodium hydroxide will decrease by the same amount.
To calculate the final concentration of acetic acid and sodium hydroxide, you need to consider the volume of the solution after mixing. In this case, you have mixed 200.0 ml of acetic acid with 200.0 ml of sodium hydroxide, resulting in a total volume of 400.0 ml (0.400 L).
Using the stoichiometry and the volume of the solution, you can determine the concentration of the remaining reactants:
Concentration of acetic acid after reaction = (moles of acetic acid / total volume)
Concentration of sodium hydroxide after reaction = (moles of sodium hydroxide / total volume)
Since the stoichiometry is 1:1, the concentration of sodium acetate will be the same as the final concentration of acetic acid and sodium hydroxide.
With the concentration of acetic acid (now sodium acetate) and the given value of Ka (1.76 x 10^-5), you can calculate the concentration of hydrogen ions (H+) and, subsequently, the pH.
pH is given by the formula: pH = -log[H+]
[H+] can be calculated using the equation for the dissociation of acetic acid:
Ka = [H+][Na(NH3COO)] / [NH3COOH]
Rearranging the equation, you get:
[H+] = sqrt(Ka x [NH3COOH] / [Na(NH3COO)])
Finally, you can use the concentration of hydrogen ions ([H+]) to calculate the pH using the formula mentioned earlier.
It is important to note that this calculation assumes the reaction goes to completion. In reality, there may be some residual acetic acid and sodium hydroxide present.
use the h-h equation : pH=pKa+log(base/acid)
since you're given ka,calculate kb using the formula:
kw/ka=kb
1*10^-14/1.76*10-5=5.68*10^-10.
pKb will therefore be -log(5.68*10^-10)=9.25
pH=9.25+log(0.5/0.5)=9.25