A 75.0 mL sample of 0.020 M acetic acid (HC2H3O2) is titrated with 0.020 M NaOH. ? Determine the pH of the solution before the addition of any NaOH. (Ka of acetic acid is 1.8 x 10-5) HC2H3O2 (aq) + H2O (l) D C2H3O2-(aq) + H3O+ (aq) (Hint: before titration so acid only, use ICE table)

pH = 3.23

Explanation:

Before the addition of any NaOH, the only you have is a 0.020M acetic acid solution. That is in equilibrium with water as follows:

HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)

The Ka of this reaction is:

Ka = 1.8x10⁻⁵ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]

Where [] are concentrations in equilibrium of each species

As you have in solution just HC₂H₃O₂, the equilibrium concentrations will be:

[HC₂H₃O₂] = 0.020M - X

[C₂H₃O₂⁻] = X

[H₃O⁺] = X

Where X is reaction coordinate.

Repalcing in Ka expression:

1.8x10⁻⁵ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]

1.8x10⁻⁵ = [X] [X] / [0.020M - X]

3.6x10⁻⁷ - 1.8x10⁻⁵X = X²

3.6x10⁻⁷ - 1.8x10⁻⁵X - X² = 0

Solving for X:

X = -0.0006M → False solution. There is no negative concentrations

X = 0.000591M → Right solution.

As:

[H₃O⁺] = X

[H₃O⁺] = 0.000591M

As pH = -log[H₃O⁺]

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.