What is the ph of a 0.0125 m naoh solution

These are two questions and two answers.

Answer:

• Question 1:   the pH of the solution after 50.0 ml of base has been added is  2.12.

• Question 2: the pH os the solution at the equivalence point is 7.

Explanation:

Question 1: What is the pH of the solution after 50.0ml of base has been added?

1) Titration: is the use of a neutralization reaction between an acid and a base to determine the concentraion of one of the solutions.

2) Neutralization reaction:

The neutralization of a strong acid and a strong base is represented by:

• Acid + Base → Salt + Water

3) HCl (strong acid) + NaOH (strong base)

• HCl + NaOH → NaCl + H₂O

4) Number of moles of each reactant:

a) 100.0 ml of 0.200 M HCl

• M = n / V (in liter) ⇒ n = M × V = 0.100 liter × 0.200 M = 0.0200 mol

b) 50.0 ml of 0.250 M NaOH

• M = n / V (in liter) ⇒ n = M × V = 0.0500 liter × 0.250 M = 0.0125 mol

5) Limiting and excess reactants:

Since, the stoichiometry is 1 : 1, 0.0125 mol of NaOH will react with 0.0125 mol of HCl. This means that NaOH is the limiting reactant, and there will be an excess of 0.0200 mol - 0.0125 mol = 0.0075 mol of HCl.

6) pH

• pH = - log [H⁺]

[H⁺] = 0.0075 mol (the same concentration of HCl left as excess because each mole of HCl ionizes into one mole of H⁺ and one mol of Cl⁻)

• pH = - log (0.0075) =  2.12

Answer: the pH of the solution after 50.0 ml of base has been added is  2.12.

Question 2.What is the pH of the solution at the equivalence point?

This is a theortical question.

The equivalence point of the titration of a strong acid with a strong base is the point where the number of moles of H⁺ from the acid equals the number of moles of OH⁻ base.

Hence, at the equivalence point, all the H⁺ and OH⁻ will be converted into H₂O and so the pH is the same pH of pure water.

Such pH is known as neutral pH and is equal to 7 (at 25°C)

So, the answer is that the pH of a the solution at the equivalence point is 7.