1. Write the chemical equation for the ionization of acetic acid. 2. From the pH

Question

1. Write the chemical equation for the ionization of acetic acid.

2. From the pH, calculate the concentration of hydrogen ions. Next, find the concentration of the acetate ion. How would the concentration of hydrogen ion be related to the acetate ion? (Hint: look at the formula for acetic acid.) Considering that acetic acid is the source of hydrogen and acetate ions in this sample, determine the actual concentration of acetic acid that is in the solution at the time when you measured the pH . Show a set of sample calculations. You might develop a talbe like this one:

0.10 M acetic acid, pH= 3.51, [H+] = ?, [C2H3O2-]=?, [HC2H3O2]=?

0.010 M acetic acid, pH= 4.30, [H+] = ?, [C2H3O2-]=?, [HC2H3O2]=?

0.0010 M acetic acid, pH= 5.75, [H+] = ?, [C2H3O2-]=?, [HC2H3O2]=?

0.0001 M acetic acid, pH= 6.28, [H+] = ?, [C2H3O2-]=?, [HC2H3O2]=?

Explanation / Answer

1. CH3COOH(aq) + H2O(l) <-----> CH3COO-(aq) + H3O+(aq)

Acetic acid Acetate ion Hydronium ion

H3O+ is equal to H+

2.

a) pH = - log [H+]

- log [ H+] = 3.51

[ H+] = 3.09 × 10^-4M

[H+] = [ CH3COO-]

Therefore, [ CH3COO-] = 3.09 × 10^-4M

Initial [CH3COOH] = 0.1M , from this concentration 3.09× 10^-4M acetic acid is dissociated to give 3.09× 10^-4M H+ ion and 3.09× 10^-4M Sodium acetate ion

Therefore, at equilibrium [CH3COOH] = 0.1M - 0.000309M = 9.969× 10^-2MM

b) pH = 4.30 Initial [CH3COOH] = 0.01M

[H+] = 5.012 × 10^-5M

[ CH3COO-] = 5.012 × 10^-5M

equilibrium [ CH3COOH] = 9.949 × 10^-3M

c) pH =5.75 initial [ CH3COOH] = 0.0010M

[H+] = 1.778 × 10^-6 M

[CH3COO-] = 1.778 × 10^-6M

equilibrium [ CH3COOH] = 9.982 ×10^-4MM

d)pH = 6.28 initial [ CH3COOH ] = 0.0001M

[ H+ ] = 5.248 × 10^-7M

[CH3COO-]= 5.248 × 10^-7M

equilibrium [ CH3COOH ] = 9.947

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