calculate pH of 0.1M CH3COOH of 100ml of that solution?
You need Ka (dissociation constant) for
CH3COOH.
then what must i do after i have the Ka of CH3COOH
Ka*[CH3COOH]/[A-]=[H3O+]
pH=-log[H3O+]
that's it :)
6.99
To calculate the pH of a solution of a weak acid like acetic acid (CH3COOH), you will need to use the equilibrium expression and the acid dissociation constant (Ka) of the acid.
1. Write the balanced equation for the dissociation of acetic acid:
CH3COOH ⇌ CH3COO- + H+
2. Write the equilibrium expression for the dissociation:
Ka = [CH3COO-] x [H+] / [CH3COOH]
3. Determine the concentration of acetic acid (CH3COOH) in moles per liter:
0.1 M CH3COOH = 0.1 moles / 0.1 L = 1 mole/L or 1 M
4. Set up an ICE (Initial, Change, Equilibrium) table to find the concentrations at equilibrium:
CH3COOH ⇌ CH3COO- + H+
I 1 M 0 M 0 M
C -x +x +x
E 1-x x x
5. Substitute the equilibrium concentrations into the equilibrium expression:
Ka = x^2 / (1-x)
6. If x is small compared to 1 (typically true for weak acids), we can assume that (1-x) will be approximately equal to 1:
Ka = x^2 / 1
7. Solve for x (concentration of H+ ions) by rearranging the equation:
x = √(Ka x 1)
8. Finally, calculate the pH using the equation:
pH = -log[H+]
Now you have all the information needed to calculate the pH of the solution. Remember to use the value of Ka for acetic acid, which is approximately 1.8 x 10^-5.