Calculate the pH after mixing 15 mL of 0.12 M acetic acid with 15 ml of 0.12 M NaOH. What are the major species in solution at equilibrium (besides water), and what are their concentrations?

Interpretation:

The pH of a solution has to be calculated when mixing 15 mL of 0.12 M NaOH with 15 mL of 0.12 M CH3COOH and The major species besides water in the solution at equilibrium and their concentration is to be stated.

Concept introduction:

The pH of a solution of acetic acid can be calculated by using the hydronium ion concentration by using the expression, pH=−log[H3O+].

Hydronium ion concentration is calculated by considering the equilibrium conditions and from the value of acid dissociation constant.

The concentration of acetic acid and acetate ion is determined by using equilibrium condition The concentration of acetate ion and hydronium ion is equal at equilibrium according to the reaction stoichiometry.

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is an acid constant for equilibrium reactions.

HA + H2O⇌H3O++ A-Ka= [H3O+][A-][HA]

Kb is a base constant for equilibrium reaction.

BOH + H2O⇌B++ OH-Ka= [B+][OH-][BOH]

The pOH is a measure of hydroxide ion concentration. The expression for pOH is,

pOH=−log[OH−] (1)

The sum of pH and pOH is equal to 14 at 25 °C.

pH+pOH=14 (2)

The pH varies from 0 to 14 for an aqueous solution.

pH=7(Neutral)pH>7(Basic)pH<7(Acidic)

An equilibrium constant (K) is the ratio of the concentration of products and reactants raised to appropriate stoichiometric coefficient at equilibrium.

For any base B, the chemical reaction is written as,

  B(aq)+H2O(l)⇌BH(aq)+OH−(aq)

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

Kb=[BH][OH−][B] (3)

An equilibrium constant (K) with subscript b indicates that it is an equilibrium constant of the base in water.

Given:

Refer to the table 16.2 for the value of dissociation constant.

Dissociation constant (Kb) of acetate ion is 5.6×10−10.

The concentration of CH3COOH is 0.12 mol⋅L−1.

The concentration of NaOH is 0.12 mol⋅L−1.

The volume of CH3COOH is 0.015​ L.

The volume of NaOH is 0.015​ L.

The balanced chemical equation involving acetic acid and sodium hydroxide is written as,

  CH3COOH(aq)+NaOH(aq)⇌CH3COONa(aq)+H2O(l)

Amount of CH3COOH consumed=(0.015 L)(0.12 mol⋅L−1)=18×10−4mol

Amount of NaOH consumed=(0.015 L)(0.12 mol⋅L−1)=18×10−4 mol 

Amount of CH3COO− produced upon completion of the reaction =18×10−4 mol

After completion of the reaction, the total volume of solution =0.030 L

The concentration of acetate ion is,

 [CH3COO−]=18×10−4mol0.030 L=0.06 mol⋅L−1=0.06 M

Therefore, the concentration of acetate ion is 0.06​ M.

Amount of Na+ produced upon completion of the reaction =18×10−4 mol

The concentration of sodium ion is,

 [Na+]=18×10−4mol0