Question

a soltion of 1.00l contains 1.0m nac2h3o2 and 1.3m hc2h3o2. the ka of acetic acid is 1.8*10^-5. what will the ph be after adding 100ml of 1.0m naoh

Answer 1

Answer:

pH = 1.94

Explanation:

First, we need to calculate the initial concentration of acetic acid and sodium acetate:

Initial [HC2H3O2] = 1.3 M
Initial [NaC2H3O2] = 1.0 M

Next, we can set up an ICE table to determine the concentration of the species after the addition of 100 mL of 1.0 M NaOH:

HC2H3O2 + OH- ⇌ C2H3O2- + H2O
I 1.3 M 0 1.0 M
C -x -x +x
E 1.3 - x - x 1.0 + x

The concentration of OH- added is 0.1 L x 1.0 mol/L = 0.1 mol/L

Using the expression for Ka, we can calculate x:

Ka = [H3O+][C2H3O2-]/[HC2H3O2]
1.8 x 10^-5 = x^2/(1.3 - x)

Simplifying this equation gives:

1.8 x 10^-5 = x^2/1.3
x = 0.00155 M

The concentration of H3O+ after the addition of NaOH is therefore:

[H3O+] = 1.8 x 10^-5 / 0.00155 = 0.0116 M

Finally, we can calculate the pH:

pH = -log[H3O+] = -log(0.0116) = 1.94

Therefore, the pH after the addition of 100 mL of 1.0 M NaOH to the solution will be 1.94.